Answer:
1. CaO is not an organic compound because it doesn’t contain a carbon molecule.
2.
Name General Structure Properties/Uses
Alcohol R-OH (contains a hydroxyl group) Can be poisonous, can be made from fermentation or distillation
Aldehyde R-COH (contains a carbon atom double-bonded to an oxygen and single-bonded to a hydrogen) Makes up formaldehyde and acetaldehyde
Ketone R-CO-R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds) Makes up acetone
Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters
Ether R-O-R (contains double carbon chains connected to an oxygen atom through single bonds) Ethyl ether is very volatile and flammable, used in veterinary medicine
3. Carbon is able to make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.
Explanation:
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CaO is not an organic compound because it doesn’t contain a carbon molecule.
Name General Structure Properties/Uses(which contains a hydroxyl group) Can be poisonous, can be made from fermentation or distillation
Aldehyde R-COH (contains a carbon atom double-bonded to oxygen and single-bonded to hydrogen) Makes up formaldehyde and acetaldehyde
Ketone R-CO-1R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds) Makes up acetone
Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters11
Ether -O-R (contains double carbon chains connected to an oxygen atom through single bonds) Ethyl ether is very volatile and flammable, used in veterinary medicine
Carbon can make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.
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Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
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The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
Which of the following represents the velocity time relationship for a falling apple?
Answer "a" would be correct.
Answer:
d
Explanation:
There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D
Brainliest please~
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.
Answer:
I = 0.25 A
Explanation:
Given that,
Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.
In series combination, the equivalent resistance is given by :
[tex]R=R_1+R_2+R_3+....[/tex]
So,
[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]
The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.
V = IR
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]
So, the current of 0.25 A passes through each bulb.
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results
Answer:
No, it will not affect the results.
Explanation:
For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.
What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.
A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
water contracts on freezing is it incorrect or conrrect
Answer:
hope it helps
much as you can
A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.
Answer:
1568J
Explanation:
Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.
Use conservation of Energy
ΔUg+ΔKE=0
ΔUg= mgΔh=2*9.8*(20-100)=-1568J
ΔKE-1568J=0
ΔKE=1568J
since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.
I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)
Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.
The 2.0-kg block feels
• the downward pull of its own weight, (2.0 kg) g
• the upward normal force of the surface, magnitude n₁
• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction
• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction
• the applied force, mag. F, pointing in the positive horizontal direction
Meanwhile the 3.0-kg block feels
• its own weight, (3.0 kg) g, pointing downward
• normal force, mag. n₂, pointing upward
• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction
• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)
Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:
• net vertical force:
n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g
• net horizontal force:
c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 35 cm while traveling through air. What are the
(a) frequency and
(b) wavelength as the signal travels through 3-mm-thick window glass into your room?
Answer:
(a) 8.57 x 10^8 Hz
(b) 23.3 cm
Explanation:
Wavelength = 35 cm = 0.35 m
speed =3 x10^8 m/s
Let the frequency is f.
(a) The relation is
speed = frequency x wavelength
3 x 10^8 = 0.35 x f
f = 8.57 x 10^8 Hz
(b) refractive index of glass is 1.5
The relation for the refractive index and the wavelength is
wavelength in glass= wavelength in air/ refractive index.
Wavelength in glass= 35/1.5 = 23.3 cm
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
what is time taken by radio wave to go and return back from communication satellite to earth??
Answer:
Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.
Explanation:
hope it help
Which phase of matter makes up stars?
O liquid
O gas
O plasma
Answer:
The answer to this question is plasma
Answer:
Plasma
Explanation:
what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
hope it is helpful to you
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.
Answer:
[tex]\frac{r_1}{r_2}=6.9[/tex]
Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:
[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]
where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,
[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]
A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.
Answer:
82.25 moles of He
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 10 L
Mass of He = 0.329 Kg
Temperature (T) = 28.0 °C
Molar mass of He = 4 g/mol
Mole of He =?
Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
0.329 Kg = 0.329 Kg × 1000 g / 1 Kg
0.329 Kg = 329 g
Thus, 0.329 Kg is equivalent to 329 g.
Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:
Mass of He = 329 g
Molar mass of He = 4 g/mol
Mole of He =?
Mole = mass / molar mass
Mole of He = 329 / 4
Mole of He = 82.25 moles
Therefore, there are 82.25 moles of He in the tank.
How do you find the product of gamma decay?
Answer:
The mass and atomic numbers don't change
Explanation:
An excited atom relaxes to the ground state emitting a photon...called a gamma ray.
The answer is that the mass and atomic numbers don't change.
In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.
To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.
During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.
The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).
For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.
It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).
Thus, the product nucleus remains unchanged in terms of atomic number and mass number.
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Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle
Answer:
Wheel and axle
Explanation:
Which simple machine is shown in the diagram?
a wheel and axle
From the given diagram, the machine shown is actually a wheel and axle
Description of wheel and axle
The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Wheel and axle
Explanation:
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?
Answer:
Explanation:
From the information given:
The angular frequency ω = 430 π rad/s
The wavenumber k = 4.00π which can be expressed by the equation:
k = ω/v
∴
4.00 = 430 /v
v = 430/4.00
v = 107.5 m/s
Similarly: k = ω/v = 2πf/fλ
We can say that:
k = 2π/λ
4.00 π = 2π/λ
wavelength λ = 2π/4.00 π
wavelength λ = 0.5 m
frequency of the wave can now be calculated by using the formula:
f = v/λ
f = 107.5/0.5
f = 215 Hz
Also, the Period(T) = 1/215 secs
The time at which particle proceeds from point A to its maximum upward displacement and to its maximum downward displacement can be computed as t = T/2;
Thus, the distance(x) covered by each wave during this time interval(T/2) will be:
x = v * t
x = v * T/2
x = λ/2
x = 0.5/2
x = 0.25 m
1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)
Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).
At point A, the block has total energy
E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²
E (A) = 686 J + 1/2 (10.0 kg) v₀²
At point B, the block's potential energy is converted into kinetic energy, so that its total energy is
E (B) = 1/2 (10.0 kg) v₁²
The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,
E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J
Throughout this whole process, energy is conserved, so
E (A) = E (B) = E (C) = E (D)
(a) Solve for v₀ :
686 J + 1/2 (10.0 kg) v₀² = 2548 J
==> v₀ ≈ 19.3 m/s
(b) Solve for v₁ :
1/2 (10.0 kg) v₁² = 2548 J
==> v₁ ≈ 22.6 m/s
Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:
• net horizontal force:
∑ F = -f = ma
• net vertical force:
∑ F = n - mg = 0
where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :
n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N
f = µn = 0.500 (98.0 N) = 49.0 N
==> - (49.0 N) = (10.0 kg) a
==> a = - 4.90 m/s²
The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that
v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)
==> v₂² = 490 m²/s²
and thus the block has total/kinetic energy
E (C) = 1/2 (10.0 kg) v₂² = 2450 J
(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so
2450 J = (10.0 kg) (9.80 m/s²) h
==> h = 25.0 m
(d) At half the maximum height, the block has speed v₃ such that
2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²
==> v₃ ≈ 15.7 m/s
The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by
v = v₁ + at = 22.6 m/s - (4.90 m/s²) t
The block comes to a rest when v = 0 :
0 = 22.6 m/s - (4.90 m/s²) t
==> t ≈ 4.61 s
It covers a distance x after time t of
x = v₁t + 1/2 at ²
so when it comes to a complete stop, it will have moved a distance of
x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m
(e) The block crosses the rough region
(52.0 m) / (2.00 m) = 26 times
vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?
Answer:
45 × 8 units = A + B as formular
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips
Answer:
I could not find the answer or do it myself if I did find it I would defenetly share
