Answer:
0.055 * 5.1 = 0.2805
Explanation:
I used a calculator so it should be accurate! yw!! :D
If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car (Assume the bottom of the cliff is zero)
Group of answer choices
the gravitational potential energy is decreasing
the gravitational potential energy has not changed
the gravitational potential energy is increasing
Explanation:
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.
so the answer is "the gravitational potential energy is decreasing"
PLEASE HELP
this is about volleyball
A good training partner helps with
• Setting the weight
• Using proper technique
• Keeping track of repetitions
• All of the above
Answer:
• All Of The Above
Hope it helps uh
Mark me as brainliest
En el proceso de diseño de ingeniería, ¿qué limitaciones deben tenerse en cuenta al utilizar un modelo / prototipo?
Answer:
Las limitaciones de un modelo o prototipo son;
1) Los parámetros ambientales (donde se opera el modelo, prototipo o producto) son diferentes y, por lo tanto, pueden producir relaciones y factores ambientales que serán diferentes de los factores ambientales y las relaciones del objeto real.
2) El análisis del problema puede ser inadecuado
3) La posibilidad de falta de satisfacción del cliente con un modelo, preferencia por la demostración real del producto.
4) Reproducción inexacta del entorno del producto durante la prueba del modelo
5) El factor de costo del modelo
6) Mayor complejidad introducida por el modelo / prototipo al análisis de la solución
Explanation:
El modelo o prototipo es la presentación del diseño articulado, construido para demostrar el producto real con el propósito de encontrar la existencia de errores en el diseño que serían corregidos, antes de que se realice la producción real
Select the correct answer. Based on the law of conservation of energy, which statement is correct? A. Energy is always being added to all parts of the Universe. B. Energy is often destroyed in some parts of the Universe. C. Energy in a closed system cannot change forms. D. Energy in an isolated system remains constant.
Answer:
D. Energy in an isolated system remains constant.
Explanation:
Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;
a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.
b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.
Furthermore, the various forms of energy are solar energy, electrical energy, chemical energy, thermal energy, wind energy, nuclear energy etc.
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
According to the law of conservation of energy, we can infer or deduce that the total energy of the system must stay the same because it cannot be destroyed, but kinetic energy (KE) can be transformed to potential energy (PE) and potential energy (PE) can be transformed to kinetic energy (KE) within the system.
1o4 Fahrenheit is equal to
Answer:
40 degrees Celsius
Explanation:
Have a great summer :)
When you flip a penny (2.35 g), it leaves your hand and moves upward at 2.85 m/s. Use energy to find how high the penny goes above your hand before stopping. A (b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor. A (c) Explain your choice of reference level for parts (a) and (b). C (d) Choose a different reference level and repeat part (b)
Answer:
a. 0.41 m
b. 5.72 m/s
c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.
ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.
d. 5.72 m/s
Explanation:
a. Use energy to find how high the penny goes above your hand before stopping.
Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.
E = E'
U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)
So, U + K = U' + K'
mgh + 1/2mv² = mgh' + 1/2mv'²
substituting the values of the variables into the equation, we have
mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²
0 + 1/2m(8.1225 m²/s²) = mgh' + 0
m(4.06125 m²/s²) = mgh'
h' = 4.06125 m²/s² ÷ g
h' = 4.06125 m²/s² ÷ 9.8 m/s²
h' = 0.41 m
(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.
Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E' equals the total mechanical energy on the ground, E"
E' = E"
U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,
So, U' + K' = U' + K'
mgh" + 1/2mv'² = mgh₁ + 1/2mv"²
substituting the values of the variables into the equation, we have
mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²
1.67mg + 0 = 0 + 1/2mv"²
1.67mg = 1/2mv"²
1.67g = 1/2v"²
v"² = 2(1.67g)
v" = √[2(1.67g)]
v" = √[2(1.67 m × 9.8 m/s²)]
v" = √[2(16.366 m²/s²)]
v" = √[32.732 m²/s²)]
v" = 5.72 m/s
(c) Explain your choice of reference level for parts (a) and (b).
i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.
ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.
(d) Choose a different reference level and repeat part (b)
Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E' equals the total mechanical energy on the ground, E"
E' = E"
U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,
So, U' + K' = U' + K'
mgh' + 1/2mv'² = mgh₂ + 1/2mv"²
substituting the values of the variables into the equation, we have
mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²
0.41mg + 0 = -1.26 mg + 1/2mv"²
0.41mg + 1.26mg = 1/2mv"²
1.67mg = 1/2mv"²
1.67g = 1/2v"²
v"² = 2(1.67g)
v" = √[2(1.67g)]
v" = √[2(1.67 m × 9.8 m/s²)]
v" = √[2(16.366 m²/s²)]
v" = √[32.732 m²/s²)]
v" = 5.72 m/s
Does understanding Earth’s place in the universe and the relationships of different objects in the solar system help people plan for the future of our planet?
Answer:
Yes
Explanation:
If there were to be a supernova of a star or a planet/meter that was directed at earth, then we would know in advance and make a plan to stop it
Which of the following describes plastic? HELP
A. A material that as a low resistivity and prevents charges from moving freely
B A material that has a high resistivity and allows charges to move freely
C A material that has a low resistivity and allows charges to move freely
B A material that has a high resistivity and prevents charges from moving freely
Answer:
D
Explanation:
A material that has a high resistivity and prevents charges from moving freely... it is part of insulators...
Plastic is a material that has a high resistivity and prevents charges from moving freely. Option D is correct.
What is plastic?Plastic is described as a material that comprises an organic compound with a significant molecular weight as an important element.
Insulators are often defined as materials that do not allow electricity to flow through them.
Insulators are also referred to as poor electrical conductors. Insulators include materials such as paper, glass, rubber, and plastic.
Plastic is a material that has a high resistivity and prevents charges from moving freely.
Hence option D is correct.
To learn more about the plastic refer to the link;
https://brainly.com/question/23509463
Please help with this Physics problem!
Answer:
Explanation:
The equation for Coulomb's Law is
[tex]F=\frac{kq_1q_2}{r^2}[/tex] where k is Coulomb's constant, q1 is one of the charges, q2 is another one of the charges (1 of these has to be the one in question, so we will let that be q2) and r is the distance between them squared.
First thing we are going to do is convert those microCoulombs to Coulombs (C).
q1: [tex]-53.0\mu C*\frac{.000001C}{1\mu C}=-5.3*10^{-5}C[/tex] and
q2: [tex]105\mu C*\frac{.000001C}{1\mu C}=1.05*10^{-4}C[/tex] and
q3: our main particle that we will put in for q2 in the formula converts as follows:
q3: [tex]-88.0\mu C*\frac{.000001C}{1\mu C} =-8.8*10^{-5}C[/tex]
First we will find the charge between q1 and the main particle:
[tex]F_1=\frac{(9.0*10^9)(5.3*10^{-5})(8.8*10^{-5})}{(1.45)^2}[/tex] Notice that we did not use the negative charges here. We take the negative charge into account depending upon whether or not the charges are repelled or attracted. Both of these charges are negative, so they will repel and the answer will be made negative. Finding the first force:
[tex]F_1=-2.0*10^1N[/tex] (negative because they repel so q1 will move away from the charge in question, which is also negative)
[tex]F_2=\frac{(9.0*10^9)(1.05*10^{-4})(8.8*10^{-5})}{(.95)^2}[/tex] and the charge between these is
[tex]F_2=92N[/tex] and that is to the right, so positive. These charges are opposite, so they attract. The net force is the sum of the forces, so:
[tex]F_1+F_2:[/tex] -2.0 × 10¹ + 92 = 72N (to the right)
Describe the difference between a sidereal day and a solar day.
Answer:
a sidereal day is the time it takes for the earth to rotate about it axis so that the distant star appears In the same position in the sky while a solar day is the time it takes for the earth to rotate about it axis so that the sun appears in the same position in the sky
A solar panel is used to recharge a battery. The solar panel produces 0.80 W of electrical power. The panel is 20% efficient. What is the power input of the sunlight onto the solar panel? * 1 point 0.16 W 4.0 W 8.0 W 16.0 W
Answer:
Input power, Ip = 4 Watts
Explanation:
Given the following data;
Power output = 0.8 Watts
Efficiency = 20%
To find the power input of the sunlight onto the solar panel;
Mathematically, the efficiency of a machine is given by the formula;
[tex] Efficiency = \frac {Out-put \; power}{In-put \; power} * 100 [/tex]
Substituting into the formula, we have;
[tex] 20 = \frac {0.8}{Ip} * 100 [/tex]
[tex] 20 = \frac {80}{Ip} [/tex]
Cross-multiplying, we have;
[tex] 20Ip = 80 [/tex]
[tex] Ip = \frac {80}{20} [/tex]
Input power, Ip = 4 Watts
an object that is 15 cm tall is placed 44 cm in front of a diverging lens. A virtual image appears -14cm in front of the lens. determine the focal length of the lens .
SHOW YOUR WORK.
A) 0.09cm
B)-20.53cm
C)10.6cm
D)-0.04cm
Answer:
The correct option is (b).
Explanation:
Given that,
The height of the object, h = 15 cm
Object distance, u = -44 cm
Image distance, v = -14 cm
We need to find the focal length of the lens. Using the lens formula.
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{f}=\dfrac{1}{(-14)}-\dfrac{1}{(-44)}\\\\f=-20.53\ cm[/tex]
So, the focal length of the lens (-20.53 cm).
Which will a positively charged object attract
The valid digits in a measurement are called _____ digits.
Answer:
Significant
Explanation:
As the word suggests, significant figures or digits are numbers that are valid in measurement.
Answer:
SIGNIFICANT DIGITS
Explanation:
MARK ME BRAINLIEST PLZZZ
Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is
A.1
B.2
C.3
D.4
(Ans with explanation pls)
Answer:
4 capacitors
Explanation:
Given
[tex]n = 5[/tex] --- conducting plates
Required
The number of capacitor (c)
This is calculated as:
[tex]c = n - 1[/tex]
So, we have:
[tex]c = 5 - 1[/tex]
[tex]c = 4[/tex]
Why does Quito, Equator has very little changes to the daylight hours
throughout the year?
Because of the position on the equator, the change in rotation of the Earth on its axis throughout the year doesn't affect it much. Unlike the poles, Quito is almost constantly in direct view of the sun. So, because of lack of change in rotation, the daylight hours are hardly varied as Quito is almost constantly in more or less the same spot in relation to the sun.
uses of convex mirrors
Explanation:
Two uses of convex mirror are: (i) It is used as a rear view mirror in vehicles. (ii) It is used as a vigilance mirror. (iii) it is used as a reflector in street lamps.
Explanation:
Inside buildings. You might have noticed that large office buildings, stores, hospitals, and other many other buildings have convex mirrors in the corners. ...
Sunglasses. We might have used sunglasses many times. ...
Vehicle mirrors. ...
Magnifying glasses. ...
For security purposes. ...
Street light reflectors.
10.
You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is
negligible friction between your feet and the ice. A friend throws you a 0.4 kg ball that is traveling
horizontally at 10 m/s. Your mass is 70 kg. If you catch the ball, with what speed do you and the ball
move afterwards?
1.02 m/s
0.06 m/s
0.02 m/s
0.12 m/s
Answer:
Explanation:
This is a classic Law of Momentum Conservation problem. For us the equation will look like this:
[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:
[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and
4.0 = 70.4v and
v = .06 m/s
I need this now
On the planet Xenos, an astronaut observes that a 2.0 m long pendulum has a period of 2.2 s.
What is the free-fall acceleration on Xenos? *
Answer:
[tex]g=16.31\ m/s^2[/tex]
Explanation:
Given that,
The length of the pendulum, l = 2 m
The period of the pendulum, T = 2.2 s
The formula for the time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
or
[tex]T^2=4\pi ^2\dfrac{l}{g}\\\\g=\dfrac{4\pi ^2l}{T^2}\\\\g=\dfrac{4\pi ^2\times 2}{(2.2)^2}\\\\g=16.31\ m/s^2[/tex]
So, the free fall acceleration is [tex]16.31\ m/s^2[/tex].
Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds
Answer:
I am not sure if this is the answer
acceleration: 2.2m/s
Explanation:
here
initial velocity(u): 11m/s
Final velocity(v): 33m/s
time taken(t): 10 s
now
a:v-u/t
or
acceleration:final velocity-initial velocity/time taken
or
a: 33-11/10
or
a:22/10, divide it
: a=2.2m/s#
When the interval between the stimuli decreases, _______.
A. a second action potential is generated until the interval reaches the absolute refractory period
B. a second action potential is generated regardless of the stimulus and the interval
C. a second action potential is generated until the interval reaches the relative refractory period
D. a second action potential is generated as long as the stimulus is above threshold
Answer:
The correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.
Explanation:
The inter-stimulus interval (ISI) is the temporal interval between two successive stimuli, measured from the offset of the first stimulus to the commencement of the second.
A cell's refractory period is the time during which it is unable to replicate an action potential. Therefore, the absolute refractory period is the amount of time it takes for a second action potential to be initiated, regardless of how large a stimulus is applied repeatedly.
A second action potential is generated when the gap between the stimuli decreases until the interval reaches the absolute refractory period.
Therefore, the correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.
That is, when the interval between the stimuli decreases, a second action potential is generated until the interval reaches the absolute refractory period.
calculate:An elcetric lamp is rated 240V and 40W. What is the cost of running the lamp for 62 hours if the elctricity authorities charges #2.50k per KWH
Explanation:
The voltage of the lamp, V = 240 V
Power of the lamp, P = 40 W
It is running for 62 hours.
The cost of running is $2.50k per KWH
Electric power is,
P = 40×62 Wh
= 2480 Wh
P = 2.48 kWh
At the rate of $2.5 per kWh
P = $6.2
So, the cost of running is $6.2 per kWh.
Charge is distributed uniformly throughout the volume of an infinitely long solid Cylinder of radius R what is the electric field when r < Select one : O a . Zero O b . E = / 2 € d . E = pr / 2 € O e . E = / 2 €
Solution :
Let us consider the Gaussian surface that is in the form of a cylinder having a radius of r and a length of A which is [tex]$\text{coaxial with the charged cylinder}$[/tex].
The charged enclosed by the cylinder is given by,
[tex]$q=\rho V$[/tex] (here, V = [tex]$\pi r^2l$[/tex] is the volume of the cylinder)
[tex]$=\pi r^2lp$[/tex]
If [tex]$\rho$[/tex] is positive, then the electric field lines moves in the radial outward direction and is normal to Gaussian surface which is distributed uniformly.
Therefore, total flux through Gaussian cylinder is :
[tex]$\phi=EA_{cyl}$[/tex]
[tex]$=E(2\pi rl)$[/tex]
Now using Gauss' law, we get
[tex]$2\pi \epsilon_0rlE = \pi r^2lp$[/tex]
or [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]
Therefore, the electric field is [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]
Hence, option (d) is correct.
what two forces contribute to the nuclear tug of war in an atom
Answer:2 protons and 2 neutrons
Explanation:In Nuclei, There are 2 forces. 1 force is electrostatic and acts as repulsion between 2 protons. The other is force of attraction called Nuclear force between 2 neutrons.
A force of 20000N acts on the raft in the direction down
State the name given to the force shown by arrow in Fig.
Calculate the mass of the raft.
Answer:
Figure is not there
Explanation:
A runner is traveling with an initial velocity of 0.3 m/s in the positive direction accelerates at a constant rate of 0.4m/s^2 for a time of 2 seconds. What is the velocity at the end of 2 seconds?
Answer:
1.1 m/s
Explanation:
Applying,
v = u+at.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, t = time.
From the question,
Given: u = 0.3 ms, a = 0.4 m/s², t = 2 seconds
Substitute these values into equation 1
v = 0.3+0.4(2)
v = 0.3+0.8
v = 1.1 m/s
Hence the velocity at the end of 2 seconds is 1.1 m/s
A.All three bulbs will go out
B.Bulbs 2 and 3 will go out, but bulb 1 will remain lit
C. All three bulbs will go out
D. Bulb 3 will go out, but bulbs 1 and 2 remain lit
Answer:
(D)
Explanation:
When switch C is opened then, Current is not flowing across 3 So bulb 3 will go out.
But current is flowing across 1 and 2 bulb because their switch is closed
therefore bulb 1 and 2 will remain it.
Hence, option (D) will be correct.
You throw a water balloon straight up with a velocity of 13 m/s. What is its
maximum height?
O A. 4.4 m
B. 6.3 m
C. 10.7 m
D. 8.6 m
Answer:
Explanation:
[tex]h=-v^2 /2g[/tex]
[tex]with\\g = 9,8 m/s^2 or 10 m/s^2[/tex]
[tex]h= (-13)^2 / 2 * 9,8 = 8,6[/tex]
to produce a magnetic field, what does an electromagnet require?
Explanation:
hope it helps
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One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 52.1 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N