A 1. 1 kg box drops two meters from a shelf and comes to rest after 0. 021 s on the floor. What force did the box hit the floor

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Answer 1

A 1 kg box falls two metres from a shelf and lands on the ground after 0.021 seconds. The box impacted the ground with a force of around 524.7 N.

The box impacted the ground with a force of around 524.7 N.

Explanation: We can determine the force using the equation F = m * (v/t), where m is the box's mass, v is the velocity change (which is the ultimate velocity because the box starts at rest and comes to a halt), and t is the time it takes for the box to stop.

Given that the box falls 2 metres and its terminal velocity is 0 m/s, v = 2 m/s.

524.7 N is equal to F = 1.1 kg * (2 m/s / 0.021 s).

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Related Questions

A car has an intial velocity of 50 km hr after 5 h, its final velocity is 70 km hr. solve for the car acceleration

Answers

Answer:

4 km/hr^2

Explanation:

We can use the formula for acceleration:

a = (v_f - v_i) / t

where:

a = acceleration

v_f = final velocity

v_i = initial velocity

t = time taken

Substituting the given values, we get:

a = (70 km/hr - 50 km/hr) / 5 hr

a = 20 km/hr / 5 hr

a = 4 km/hr^2

you live on an island in the pacific. an earthquake of magnitude 8.5 off the coast of japan, 8000 km away, generates a tsunami with a wavelength of 200 km. the average water depth between your island and japan is 4900 m. if a tsunami warning is issued for your island, how many hours will you have before the waves arrive?

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If a tsunami warning is issued for the island, they will have approximately 11.7 hours before the waves arrive.

What is Magnitude?

Magnitude is a measure of the strength or intensity of a physical quantity or phenomenon, such as an earthquake or a sound wave. It is often expressed using a numerical scale, with higher values indicating greater strength or intensity. In the case of earthquakes, magnitude is typically measured using the Richter scale or the moment magnitude scale, which take into account the amplitude of seismic waves and the energy released by the earthquake.

To calculate the time it takes for a tsunami to travel from Japan to the island, we can use the following formula:

t = (2 * pi * d) / g * ln(1 + sqrt(h/d))

where t is the time it takes for the tsunami to travel, d is the average water depth, h is the wave height, and g is the acceleration due to gravity (9.8 m/s^2).

Magnitude of the earthquake: 8.5

Wavelength of the tsunami: 200 km = 200,000 m

Average water depth: 4,900 m

To calculate the wave height, we can use the following formula:

h = (M / 5) * (D / 10)^1/2

where M is the magnitude of the earthquake and D is the distance between the earthquake epicenter and the observation point (in this case, the island). Note that this formula is an approximation and may not be accurate for all cases.

Using the given values, we get:

D = 8,000 km = 8,000,000 m

h = (8.5 / 5) * ((8,000,000 / 10)^1/2) = 2,738.6 m

Substituting these values into the formula for t, we get:

t = (2 * pi * 4,900) / 9.8 * ln(1 + sqrt(2,738.6/4,900)) = 11.7 hours

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a particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. the particle's position at t0=0s is x0 = -5.40 m . at t1 = 2.00 s , the particle is at x1 = 5.80 m .

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A particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. The particle's position at t0=0s is x0 = -5.40 m. At t1 = 2.00 s , the particle is at x1 = 5.80 m. The value of k is 2.80 m/s2.

The given equation describes the velocity of a particle in terms of a constant, k, and time, t. The velocity, vx, is given in m/s. The initial position of the particle at t0=0s is x0=-5.40 m, and at t1=2.00 s the particle is at x1=5.80 m. To find the value of the constant k, we can solve the equation for the change in velocity Δvx.

Δvx = vx1 – vx0 = k(t12 – t02)
Δvx = 5.80 – (-5.40) = 11.20 m/s

k = (11.20 m/s) / (2.002 s2) = 2.80 m/s2

Now that we have found the value of the constant k, we can use it to find the velocity of the particle at any time t. For example, at t2=4.00 s the velocity of the particle is vx2=11.20 m/s. This can be calculated using the equation vx2 = k(t22) = 2.80(4.002) = 11.20 m/s.

From the velocity equation, we can also calculate the position of the particle at any time t. The position of the particle at t2=4.00 s is x2= 11.20(4.00) = 44.80 m. We can also calculate the position of the particle at any other time t, by simply substituting in the corresponding value of t into the equation.

In conclusion, the equation vx = kt2 describes the velocity of a particle in terms of a constant, k, and time, t. Using this equation, we can calculate the velocity and position of the particle at any given time.

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Complete Question:

A particle’s velocity is described by the function vx = [tex]kt^2m/s[/tex], where k is a constant and t is in s. The particle’s position at [tex]t_0[/tex] = 0s is [tex]x_0[/tex] = -5.40 m. At [tex]t_1[/tex] = 2.00 s, the particle is at [tex]x_1[/tex] = 5.80 m. Determine the value of the constant k. Be sure to include the proper units

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 12 ounces. a. The process standard deviation is 0.14, and the process control is set at plus or minus 2.4 standard deviations. Units with weights less than 11.664 or greater than 12.336 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found to the nearest whole number)? b. Through process design improvements, the process standard deviation can be reduced to 0.12. Assume the process control remains the same, with weights less than 11.664 or greater than 12.336 ounces being classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found to the nearest whole number)?

Answers

a. To calculate the probability of a defect, we need to find the area under the normal distribution curve that falls outside the control limits of 11.664 and 12.336 ounces. We can calculate the z-scores for these limits as follows:

[tex]z_1 = (11.664 - 12) / 0.14 = -2.4[/tex]

[tex]z_2 = (12.336 - 12) / 0.14 = 2.4[/tex]

Using a standard normal distribution table or calculator, we can find that the probability of a defect is approximately 0.0115 (to 4 decimals).

To find the expected number of defects in a production run of 1000 parts, we can use the formula for the binomial distribution:

[tex]P(X = k) = C(n, k) \times p^k \times (1-p)^{(n-k)}[/tex]

where P(X = k) is the probability of exactly k defects in a run of n parts, p is the probability of a single defect (0.0115 in this case), and C(n, k) is the binomial coefficient (the number of ways to choose k defects from n parts).

For k = 0, 1, 2, ..., we can calculate the probabilities and add them up to find the expected number of defects:

E(X) = sum(k=0 to n) [ P(X = k) ] = n * p

Substituting n = 1000 and p = 0.0115, we get:

[tex]E(X) = 1000 \times 0.0115 = 11.5[/tex]

So we can expect to find approximately 12 defects (to the nearest whole number) in a production run of 1000 parts.

b. With a reduced process standard deviation of 0.12, the z-scores for the control limits remain the same as in part a:

[tex]z_1 = (11.664 - 12) / 0.12 = -2.8[/tex]

[tex]z_2 = (12.336 - 12) / 0.12 = 2.8[/tex]

Using a standard normal distribution table or calculator, we can find that the probability of a defect is approximately 0.0004 (to 4 decimals).

To find the expected number of defects in a production run of 1000 parts, we can use the same formula as in part a:

[tex]E(X) = n \times p[/tex]

Substituting n = 1000 and p = 0.0004, we get:

[tex]E(X) = 1000 \times 0.0004 = 0.4[/tex]

So we can expect to find approximately 0 defects (to the nearest whole number) in a production run of 1000 parts.

However, it's important to note that this assumes the process is operating exactly at the mean weight of 12 ounces and there is no other source of variation. In practice, there may still be some small amount of variation that could result in a few defects.

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Is an object moving with a constany speed around a circular path veloctiy? why? why not?​

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Answer: The motion of a body with constant speed in a circular path is said to be accelerated, because it is moving with uniform speed, but not with uniform velocity, as velocity is a vector quantity, it can be represented in magnitude as well the direction.

Explanation:

A piece of metal weighing 187.6 g is placed in a graduated cylinder containing 225.2 mL of water. The combined volume of solid and liquid is 250.3 mL. What is the density, in grams per milliliter, of the metal?

Answers

The density of the metal in grams per milliliter is 7.87 g/mL.

Given data:The weight of metal, W = 187.6 g,Volume of water, V₁ = 225.2 mL.

The combined volume of solid and liquid, V₂ = 250.3 mL

Volume of the metal can be calculated as:Volume of metal = V₂ - V₁= 250.3 - 225.2= 25.1 mL

The density of the metal can be calculated as:Density = Weight of metal / Volume of metal

Density = W / V= 187.6 g / 25.1 mL= 7.87 g/mL

Thus, the density, in grams per milliliter, of the metal is therefore calculated and found to be 7.87 g/mL.

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the cardinals kick a 0.43 kg football for a 3-point field goal. if the ball is kicked at 24 m/s at an angle of 53-degrees, how far will it go before landing back on level ground?

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The distance which the football which cover before landing back on the ground level will be about 56.4 meters.

What is the distance of football?


The mass of football, m = 0.43 kg, Initial velocity of football (v) = 24 m/s, Angle of inclination(θ) = 53°

From the given data, we know that the vertical component of the initial velocity is given by, vsin(θ) and the horizontal component of initial velocity is given by, vcos(θ). So, the time taken by the football to reach the maximum height is given by,

t = (vsin(θ))/g

Here, g = 9.8 m/s²

Now, the maximum height attained by the football is given by,h = (vsin(θ))²/(2g).

Therefore, the time of flight or the total time which is taken by the football to land on the ground level is given by,

T = 2t

Now, the horizontal distance travelled by the ball is given by, d = (vcos(θ))T

Substituting the given values in the above formulas, we get:

t = (24sin(53°))/9.8 = 1.71 s

h = (24sin(53°))²/(2×9.8) = 23.4m

T = 2×1.71 = 3.42 s

d = (24×cos(53°))×3.42 = 56.4 m

Therefore, the football will go 56.4 m before it is landing back on the level ground.

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of the three states of matter, which one has the most kinetic energy?

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Of the three states of matter (solid, liquid, and gas), gas has the most kinetic energy. This is because the particles in a gas have the highest average speed compared to the particles in solids and liquids.

In a gas, the particles are in constant motion, colliding with each other and the walls of the container. This motion generates kinetic energy, which is proportional to the speed and mass of the particles. In contrast, solids have the lowest kinetic energy because their particles are tightly packed and have limited movement. The particles in a solid vibrate around a fixed position, and only experience small oscillations. Liquids have an intermediate amount of kinetic energy. The particles in a liquid are less tightly packed than in a solid, and can move more freely, resulting in more kinetic energy. However, liquids have more intermolecular forces between the particles compared to gases, which restricts their movement and reduces their average speed. Therefore, of the three states of matter, gases have the most kinetic energy, followed by liquids and then solids.

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help me
plss asap!!!

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Answer:B

Explanation:The ray above makes a 90 degree angle. The ray below makes a 60 degree angle.

the end result of a theory that is not verified is

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Unproven theories ultimately cannot be regarded as scientific facts or principles and are not generally recognised by the scientific community.

A well-supported explanation of a natural occurrence in science that has passed rigorous examination and is backed by empirical data is referred to as a theory. A hypothesis, however, cannot be regarded as a scientific fact or principle if it is not backed up by empirical data or if it has not undergone extensive testing and verification. The scientific community frequently rejects unproven notions with scant empirical backing and may even label them as pseudoscientific or non-scientific. This is so that scientific theories and findings may be evaluated and verified frequently. Science does this by using evidence-based reasoning and critical thinking. Unproven theories are therefore eventually not regarded as being a part of the corpus of scientific knowledge.

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Ceteris paribus, which of these events would cause both the equilibrium interest rate and the equilibrium quantity of investment to fall?
• A dcrease investor confidance
• A decrease in cosmetic income and wealth • A strengh of time preference
• A decrease in capital productivity

Answers

Ceteris paribus, a decrease in capital productivity is the event that would cause both the equilibrium interest rate and the equilibrium quantity of investment to fall. The correct answer is option C.

Ceteris paribus is a Latin expression that means "all other things being equal." Ceteris paribus is a model in which economists use to analyze the effect of one independent variable on a dependent variable while keeping all other independent variables constant. This implies that only one variable is allowed to change while all other variables are held constant at their current level or position.

Therefore, Ceteris paribus, an increase in investor confidence, an increase in cosmetic income and wealth, and a strength of time preference will not cause both the equilibrium interest rate and the equilibrium quantity of investment to fall. However, a decrease in capital productivity is an event that would cause both the equilibrium interest rate and the equilibrium quantity of investment to fall.

When capital productivity is low, firms are unable to produce goods and services efficiently, and as a result, the demand for investment falls. When the demand for investment falls, the equilibrium quantity of investment will also decrease, leading to a decrease in the equilibrium interest rate.

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Categorize the following exercises as being isometric or isotonic.
Pushing constantly against a concrete wall
Running up a hill
Swimming freestyle
Pedaling a bicycle on a flat surface
Holding a bench-press bar in the same position
Doing a plank exercise (holding a push-up position)
Balancing on tiptoes
Doing bicep curls

Answers

Isometric pushes against a wall made of concrete, Isotonic running up a hill. isotonic freestyle swimming, bicycle pedalling on a level surface: isotonic.

Static muscle contractions, in which the length of the muscle does not change during the workout, are called isometric exercises. This indicates that during the activity, there is no discernible movement or alteration in joint angle. Instead, the muscles are tense against a constant force or maintained still for a certain period of time. Exercises that are isometric include pushing against a wall, keeping a plank position, and tightening a hand grasp. Exercises that are isometric can help to increase joint stability and balance as well as muscular strength and endurance. They can also be incorporated into normal workout routines for general health and strength training. They are frequently used in physical therapy to aid patients in recovering from injuries or surgery.

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for the given input voltage amplitude (200 mvpp), what is the maximum gain that this amplifier will be able to produce? show your calculation below.

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The maximum gain of an amplifier that produces an output voltage amplitude of 50 Vpp with an input voltage amplitude of 200 mVpp is 25. The formula to calculate gain is output voltage amplitude divided by input voltage amplitude.

In this case, we are given an input voltage of 200 mVpp, so the maximum gain of this amplifier can be calculated as follows:

Gain = Output Voltage/Input Voltage = Output Voltage/200 mVpp

Therefore, the maximum gain of this amplifier is equal to the output voltage. In other words, the maximum gain of this amplifier is equal to the voltage output of the amplifier.

To calculate the output voltage of the amplifier, we need to know the supply voltage and the resistance of the load. Assuming the supply voltage is 5V and the load resistance is 10k ohms, the output voltage can be calculated as follows:

Output Voltage = Supply Voltage * Load Resistance / (Load Resistance + Output Resistance) = 5V * 10k ohms / (10k ohms + 10k ohms) = 5V

Therefore, the maximum gain of this amplifier is 5V/200 mVpp = 25.

To summarize, the maximum gain of this amplifier is 25, calculated by dividing the output voltage by the input voltage. The output voltage can be calculated by knowing the supply voltage and load resistance.

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if two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? explain

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The battery has to supply more power when two resistors are connected in series than when only one resistor is connected. This is because the power dissipated in a series circuit is equal to the sum of the power dissipated in each resistor.


When two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected. This is because the resistors offer resistance, which results in the dissipation of energy as heat. The higher the resistance of a resistor, the more power it requires to operate.Resistors consume energy as they offer resistance to the flow of current. The power supplied by the battery is converted to heat energy in the resistor, and the amount of heat energy dissipated is determined by the resistance of the resistor. The greater the resistance of the resistor, the more power it requires to function.

As a result, when two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected, to produce the same current through the circuit. Therefore, if two resistors of equal value are connected in series, the total power dissipated is twice that of when a single resistor is connected.

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The magnitude of the force between two point charges 1. 0 m

apart is 9 x 10°n. If the distance between them is doubled,

what does the force become?

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Force will become  2.25 x 10^N. because, According to Coulomb's Law, the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Thus, if the distance between two point charges is doubled, the force between them will decrease by a factor of 4. This is because the inverse square relationship means that the force decreases rapidly with distance. Therefore, if the force between two point charges is 9 x 10^N when they are 1 meter apart, when the distance is doubled to 2 meters, the force will become 9 x 10^N / 4 = 2.25 x 10^N.

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What allowed the Voyager 2 spacecraft to make a "tour" of all four of the jovian planets in the late 1970's and the 1980's?
1) NASA had developed a new kind of rocket that could propel the craft from planet to planet
2) the four planets were approximately aligned on one side of the Sun and we used the gravity of each planet to speed up the spacecraft to get to the next one in its path
3) the spacecraft stopped off to collect fuel on the satellites of each planet before proceeding to the next one
4) we used laser beams to propel the spacecraft into the outer solar system, where sunlight is dim
5) you can't fool me, no single spacecraft has ever explored four different planets

Answers

Answer:

The four planets were approximately aligned on one side of the Sun and we used the gravity of each planet to speed up the spacecraft to get to the next one in its path

Explanation:

All the Options 1, 2, 3, 4  are true about the Voyager 2 spacecraft to make a "tour" of all four of the jovian planets in the late 1970's and the 1980's.

The Voyager 2 spacecraft was able to make a "tour" of all four of the jovian planets in the late 1970's and the 1980's due to the following:

NASA had developed a new kind of rocket that could propel the craft from planet to planet.The four planets were approximately aligned on one side of the Sun and we used the gravity of each planet to speed up the spacecraft to get to the next one in its path.The spacecraft stopped off to collect fuel on the satellites of each planet before proceeding to the next one.We used laser beams to propel the spacecraft into the outer solar system, where sunlight is dim.

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a researcher is studying the distribution of auxin in roots and stems exposed to sunlight. he notices that more auxin collects in the sides of stems and roots that are not exposed to light. why?

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The researcher's observation that more auxin collects in the sides of stems and roots that are not exposed to light is likely due to the phenomenon of phototropism.

In the process of phototropism, light influences the direction and rate of growth of plant cells. In particular, light induces the cells on one side of a stem or root to create less auxin than the cells on the shaded side. Less auxin is produced on the lighted side and more auxin is produced on the shaded side as a result. The hormone auxin is essential for controlling the growth and development of plants. Auxin generally promotes cell growth and elongation at greater concentrations while inhibiting cell elongation at lower concentrations. Since the cells on the lighted side of the stem or root will contain less auxin when there is light.

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three forces applied to a trunk that moves leftward by 3.010 m over a frictionless floor. The force magnitudes are F1 = 5.86 N, F2 = 9.180 N, and F3 = 3.850 N, and the indicated angle is θ = 67.8°. During the displacement, what is the net work done on the trunk by the three forces? (Note that there are other forces acting on the block, but we only care about the net work done by these three forces.) And by how much does the kinetic energy of the trunk increase (enter a positive value) or decrease (negative value)?

Answers

The kinetic energy of the trunk increases by ½ mvf² = ½ m(10.65 m/s)²= 71.44 J during the displacement.

Net work = ΔK

W = Fd cosθ

W1 = F1d cosθ = (5.86 N)(3.010 m) cos(67.8°) = 6.99 J

W2 = F2d cosθ = (9.180 N)(3.010 m) cos(67.8°) = 10.97 J

W3 = F3d cosθ = (3.850 N)(3.010 m) cos(67.8°) = 4.58 J

Net work = W1 + W2 + W3 = 6.99 J + 10.97 J + 4.58 J = 22.54 J

Therefore, the net work done on the trunk by the three forces is 22.54 J.

ΔK = ½ mvf² - ½ mvi²

Since the trunk moves a distance of 3.010 m and is initially at rest, we can use the equation:

vf² = 2ad

where a is the acceleration of the trunk, which is given by:

a = ΣF / m

where ΣF is the net force on the trunk, which we can find using:

ΣF = F1 + F2 + F3

ΣF = (5.86 N + 9.180 N + 3.850 N) = 18.89 N

Therefore, the acceleration of the trunk is:

a = ΣF / m = 18.89 N / m

Since the trunk moves leftward, the acceleration is also leftward, so we can use a negative value for a.

Substituting the values for a and d, we get:

vf² = -2(-18.89 N / m)(3.010 m) = 113.51 (m/s)²

Taking the square root, we get:

vf = 10.65 m/s

Therefore, the change in kinetic energy of the trunk is:

ΔK = ½ mvf² - ½ mvi² = ½ m(10.65 m/s)²- 0 = ½ mvf²

Kinetic energy is a type of energy that an object possesses by virtue of its motion. It is defined as the energy an object has due to its motion and is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is KE = 1/2mv^2, where m is the mass of the object and v is its velocity.

Kinetic energy is a scalar quantity and has units of joules in the International System of Units (SI). It is a fundamental concept in physics and is used to describe many physical phenomena, including the motion of particles, the behavior of gases, and the motion of waves. In many cases, kinetic energy can be transformed into other forms of energy. For example, when a ball is thrown upwards, its kinetic energy is gradually converted into gravitational potential energy as it moves higher and higher.

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Solve the circuit shown in the figure above, also explain how you did it

Answers

Answer:

Explanation:

Using Kirchhoff's laws, we can solve for the current i:

At the node where the 2Ω and 4Ω resistors meet, the current is split into two branches, i and i1. Applying Kirchhoff's current law (KCL), we have:

i + i1 = 12/2 = 6 A

At the loop with the 2Ω, 4Ω, and 5Ω resistors, applying Kirchhoff's voltage law (KVL), we have:

-20 + 2i + 4i1 + 5i1 = 0

-20 + 6i1 + 2i = 0

6i1 + 2i = 20

3i1 + i = 10

We can solve this system of equations by substitution, which gives:

i = 2 A

Therefore, the current through the 2Ω resistor is 2 A. The answer is (A) 2 A.

when a 2.75-kg fan, having blades 18.5 cm long, is turned off, its angular speed decreases uniformly from 10.0 rad/s to 6.30 rad/s in 5.00 s. (a) what is the magnitude of the angular acceleration of the fan?

Answers

The angular acceleration of the fan is 0.740 rad/s^2,

Angular acceleration which represents the rate at which the angular velocity changes over time. The unit used to measure angular acceleration is radians per square second (rad/s2), according to the International System of Units. The Greek alphabet symbol alpha (α) is used to denote angular acceleration.

To calculate the angular acceleration of the fan, the formula α = Δω/Δt is used. Here, α represents angular acceleration, Δω represents the change in angular speed, and Δt represents the change in time.

In this scenario, Δω is equal to 10.0 - 6.30 = 3.70 rad/s, and Δt is equal to 5.00 s. By substituting these values into the formula, we obtain α = 3.70/5.00 = 0.740 rad/s^2.

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!!! If each compound undergoes electrophilic aromatic substitution, where should the substituent be added? Phenol?
Benzaldehyde?
Benzoic Acid?
Bromobenzene?
Nitrobenzene?
Toluene?

Answers

The substituent in Phenol is added to the ortho and para positions of the benzene ring. The substituent in Benzaldehyde is added to the ortho and para positions of the benzene ring.

The substituent in Bromobenzene is added to the ortho and para positions of the benzene ring. The substituent in Nitrobenzene is added to the meta position of the benzene ring. The substituent in Toluene is added to the ortho and para positions of the benzene ring.

Substituents on different aromatic compounds. The substituent is added to different positions for each of the aromatic compounds if they undergo electrophilic aromatic substitution. The positions where the substituents are added to Phenol, Benzaldehyde, Benzoic Acid, Bromobenzene, Nitrobenzene, and Toluene are described below:

Phenol- The substituent in Phenol is added to the ortho and para positions of the benzene ring.

Benzaldehyde- The substituent in Benzaldehyde is added to the ortho and para positions of the benzene ring.

Benzoic Acid- The substituent in Benzoic acid is added to the meta position of the benzene ring.

Bromobenzene- The substituent in Bromobenzene is added to the ortho and para positions of the benzene ring.

Nitrobenzene- The substituent in Nitrobenzene is added to the meta position of the benzene ring.

Toluene- The substituent in Toluene is added to the ortho and para positions of the benzene ring.

Thus, we can see that the positions of the substituent in each aromatic compound depend on the particular compound that undergoes electrophilic aromatic substitution.

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1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ____________ 3. I’m the only metal who is a liquid at room temperature. Who am I? ____________ 4. I’m named after the person who created the 1st Periodic Table. Who am I? ___________ 5. I have 92 protons. Who am I? _____________ 6. I’m the only nonmetal who is a liquid at room temperature. Who am I? ___________ 7. I’m named after a very famous scientist. Who am I? ___________ 8. I have 46 electrons. Who am I? ____________ 9. My atomic mass is 183. 84. Who am I? _____________ 10. My chemical symbol is Ag. Who am I? ________________ 11. I’m the only metalloid in period 3. Who am I? ___________ 12. I’m the only element that is solid and a nonmetal in group 14. Who am I? _____________ 13. I have 5 neutrons. Who am I? ____________ 14. I’m the only gas at room temperature that is in group 16. Who am I? ___________ 15. I have 68 protons. Who am I? __________ 16. What element has the chemical symbol of Ir? ______________ 17. Which element is in group 7 and has 30 neutrons. Who am I? ___________ 18. I’m the only metal in group 15. Who am I? ____________ 19. I have 88 electrons. Who am I? ___________ 20. I’m the only gas at room temperature and in period 5. Who am I? ____________ 21. My symbol is Am. Who am I? ______________ 22. I’m the only nonmetal in period 6. Who am I? ____________ 23. My atomic number is 69. 723. Who am I? _________________ 24. I have 159 neutrons. Who am I? ________________ 25. I’m the only metalloid in group 17. Who am I? ______________ 26. I have 50 electrons. Who am I? __________________ 27. I’m in the 1st group and the 4th period. Who am I? ________________ 28. I’m a metalloid whose symbol is Sb. Who am I? ______________ ©JFlowers2017 Name: ______________________________ Date: ___________Class: ________ Periodic Table Scavenger Hunt Directions: You will use the Periodic Table to answer the questions. 1. I’m in the 17th column, a nonmetal, & a solid at room temperature. Who am I? ________________ 2. I have 79 electrons. Who am I? ____________ 3. I’m the only gas in period 6. Who am I? ____________ 4. My atomic mass is 257. Who am I? ___________ 5. My chemical symbol is Hs. Who am I? _____________ 6. I have 114 neutrons. Who am I? ___________ 7. I’m in the 18th group and 2 nd period. Who am I? ___________ 8. I have 67 protons. Who am I? ____________ 9. I’m a nonmetal who is solid at room temperature & has 2 letters for my symbol. Who am I? _________ 10. I’m in the 1 st group & 7 th period. Who am I? ________________ 11. I’m the only metalloid in group 13. Who am I? ___________ 12. I have 97 electrons. Who am I? _____________ 13. I am the only gas in column 15. Who am I? ____________ 14. My name is similar to Mickey Mouse’s best friend. Who am I? ___________ 15. I’m in group 11 & period 4. Who am I? __________ 16. I have 62 protons. Who am I? ______________ 17. My name fits really well with doctors because they try to do this. Who am I? ___________ 18. My name reminds me of where we all live. Who am I? ____________ 19. I’m the only nonmetal in period 2. Who am I? ___________ 20. My atomic number is 87. 62. Who am I? ____________ 21. My symbol is Mt. Who am I? ______________ 22. I’m in group 17 & the only metalloid. Who am I? ____________ 23. I have 71 electrons. Who am I? _________________ 24. My symbol is Pd. Who am I? ________________ 25. I’m Dorothy’s friend who needed a heart. Who am I? ______________ 26. I have 41 protons. Who am I? __________________ 27. I have 125 neutrons. Who am I? ________________ 28. My name comes from the 8th planet. Who am I? ______________

Answers

The Periodic Table of Elements served as the inspiration for this scavenger hunt. The exercise consists of two sets of questions, each of which has 28 questions that must be answered using the Periodic Table.

Students are tasked with identifying elements in the first set of questions using information from their attributes, such as the element's position on the periodic table, atomic mass, or quantity of electrons, protons, or neutrons. The objectives of the questions are to familiarise students with the properties of various elements and the structure of the Periodic Table. The second series of questions is comparable to the first, but more difficult because it asks students to identify components using less obvious cues, like their chemical symbol or a chemical formula. In order to succeed in their future studies of chemistry and other related sciences, students will benefit from being more familiar with the structure of the periodic table and the characteristics of various elements.

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Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20 m high. Assume it starts from rest and rolls without slipping.
Express your answer using three significant figures and include the appropriate units. Thank you!!

Answers

The translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.

We can use conservation of energy to solve this problem. The initial energy of the cylinder is all potential energy, and the final energy is all kinetic energy. The potential energy at the bottom of the incline is zero.

The potential energy of the cylinder at the top of the incline is given by:

PE = mgh

where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the incline. Substituting the given values, we get:

PE = (mass of cylinder) x (acceleration due to gravity) x (height of incline) = mgh

The kinetic energy of the cylinder at the bottom of the incline is given by:

KE = (1/2)mv^2

where v is the translational speed of the cylinder at the bottom of the incline.

According to the conservation of energy, the initial potential energy is equal to the final kinetic energy, so we can set these two expressions equal to each other:

mgh = (1/2)mv^2

We can cancel the mass of the cylinder from both sides, and solve for v:

v = sqrt(2gh)

Substituting the given values, we get:

v = sqrt(2 x 9.81 m/s^2 x 7.20 m) ≈ 9.43 m/s

Therefore, the translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.

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You are the process engineer at Corvallis Automobiles Inc., and you have received an order to turn a cylindrical bar on an engine lathe to the dimensions specified in Fig. 1. For this order you will use cylindrical bar stock that is 48-inches long and 4-inches in diameter. The 48-inch length bar will be chucked in the lathe and supported at the opposite end using a live center. You are planning to complete the operation in one pass using a cutting speed of 400 ft./min. and a feed of 0.010 in./rev. Determine the following: a) The required depth of cut (in inches) b) The material removal rate (in cubic inches per minute)
c) The time required to complete the cutting pass (in minutes)

Answers

a.  the depth of cut  is 0.625 inches.

b. the material removal rate is 0.003125 cubic inches per minute.

c. the time required to complete the cutting pass is 20 minutes.

How do we calculate?

a) The required depth of cut can be determined by :

DOC = (4 in - 2.75 in)/2 = 0.625 in

Therefore, the depth of cut is  0.625 inches.

b) The material removal rate can be found by applying:

MRR = DOC x Width of cut x Feed rate

assuming we are using a standard carbide insert tool with a width of cut of 0.5 inches.

MRR = 0.625 in x 0.5 in x 0.010 in/rev = 0.003125 cubic inches per minute

c) The time required to complete the cutting pass is determined by:

Time = Length of cut / (Cutting speed x Width of cut x Feed rate)

Time = 48 in / (400 ft/min x (0.5 in) x (0.010 in/rev) x (1/12 ft/in)) = 20 minutes

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a vhf television station assigned to channel 22 transmits its signal using radio waves with a frequency of 518 mhz. calculate the wavelength of the radio waves. round your answer to significant digits.

Answers

The wavelength of the radio waves is approximately 0.579 m or 57.9 cm

Wavelength is the distance covered by an electromagnetic wave while propagating through space. The relationship between the wavelength and the frequency of an electromagnetic wave is given by the formula;

Wavelength = speed of light / frequency = c / f

where c is the speed of light and f is the frequency of the wave.

To calculate the wavelength of a VHF television station assigned to channel 22 that transmits its signal using radio waves with a frequency of 518 MHz, we substitute the known values into the equation above.

Wavelength = c / f = (3 x 10⁸ m/s) / (518 x 10⁶ Hz) = 0.579 m or 57.9 cm (rounded to three significant digits)

Therefore, the wavelength of the radio waves transmitted by the VHF television station assigned to channel 22 is 0.579 m or 57.9 cm (rounded to three significant digits).

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In outer space will a liquid in a beaker exert a pressure on the bottom or on the sides of a beaker?

Answers

Answer:

Yo dude, if you had a beaker of liquid in outer space, it wouldn't push down on the bottom or the sides of the beaker like it would on Earth. In space, there's no gravity to make the liquid settle down, so it forms a round shape because of surface tension. So basically, the liquid would just float around in a ball inside the beaker. If you moved the beaker around, the liquid would just roll around with it like a bouncy ball.

the maximum energy of photoelectrons from aluminium is 2.3 ev for radiation of 2000 a and 0.90 ev for radiation of 3130 a. use this data to calculate plancks constant and the work function of aluminium

Answers

The maximum energy of photoelectrons from aluminium is 2.3 eV for radiation of 2000 Å and 0.90 eV for radiation of 3130 Å.

To calculate Planck's constant and the work function of aluminium, we need to use the equation:


 h = E2 - E1/ λ2 - λ1

Where h is Planck's constant, E1 and E2 are the maximum energy of photoelectrons for each wavelength, and λ1 and λ2 are the wavelengths.

Using the given data, we have:

h = (2.3 - 0.90) / (2000 - 3130)

Therefore, h = -1.4 eV / -930 Å, which simplifies to h = 0.0015 eVÅ.

The work function of aluminium is equal to the maximum energy of the photoelectrons for the longest wavelength, in this case, 0.90 eV. Therefore, the work function of aluminium is 0.90 eV.

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A 2 kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. The

object's speed after falling for 3 sis 75 m/s. Air resistance is considered to be negligible, Calculate the weight of the 2 kg object on the planet of unknown mass.

2N

B

25 N

50N

D

75 N

Answers

The Answer is 50N .

In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the ................. (A) -x direction at a constant 10 m/s. (B) - direction increasing in speed. (C) +x direction increasing in speed. (D) - direction decreasing in speed. (E) +x direction decreasing in speed.

Answers

In the case where the car is traveling in the -x direction and decreasing in speed, it has a negative velocity and a positive acceleration. Therefore, option D is the correct answer. In this case, the car is traveling in the - direction and decreasing in speed. Therefore, it has a negative velocity and a positive acceleration.

Let's discuss the given options one by one:

(A) In this case, the car is traveling in the -x direction at a constant speed. Therefore, it has a negative velocity and zero acceleration. This option is incorrect.

(B) In this case, the car is traveling in the - direction and increasing its speed. Therefore, it has a negative velocity and a positive acceleration. However, the given direction is not specified, and thus this option is not accurate.

(C) In this case, the car is traveling in the +x direction and increasing in speed. Therefore, it has a positive velocity and a positive acceleration. This option is incorrect.

(D) In this case, the car is traveling in the - direction and decreasing in speed. Therefore, it has a negative velocity and a positive acceleration. This option is correct.

(E) In this case, the car is traveling in the +x direction and decreasing in speed. Therefore, it has a positive velocity and a negative acceleration. This option is incorrect.

Therefore, Option D ( - direction decreasing in speed) is correct.

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Find the fourier series of f(x)=x
for 0<=x<=2

Answers

The function f(x) = x, where 0 x 2, has the following Fourier series: Given that f(x) has an odd period of 2, we may express its Fourier series as follows: F(x) = A0 + [n=1 to] Ancos (n/x) plus bnsin (n/x).

Since f(x) is an odd function, a0 = 0. We may apply the following formulae to determine the Fourier coefficients: a = (2/1) f(x)cos(nx/1)[0 to 1] dx Bn = (2/1) f(x)sin(nx/1)[0 to 1] dx We may determine the coefficients using the following formulas: an is equal to (2/1) [0 to 1] x*cos(nx/1) dx. Bn is equal to (2/1), [0 to 1]x*sin(nx/1)dx. By integrating in pieces, we obtain: a = (2/π^2) [(1-(-1)^n)/(n^2)] bn = (2/π) [(1-(-1)^n)/(n)] The Fourier series of f(x) = x, where 0 x  its Fourier series as follows: F(x) = A0 + [n=1 to] Ancos (n/x) plus bnsin (n/x).2, is as follows: f(x) = Σ(n=1 to ∞) [(2/) (1-(-1)n)/(n))*sin(nx/1)].

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