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Neko [114]
3 years ago
11

Form a hypothesis. Which will be greater, the potential energy of the car at the top of the ramp or the kinetic energy of the ca

r at the bottom of the ramp
Physics
1 answer:
AleksandrR [38]3 years ago
6 0

The potential energy of the car at the top of the ramp is greater than the kinetic energy of the car at the bottom of the ramp

Explanation:

Here we have a car moving down along a ramp. In absence of force of friction, the total mechanical energy of the car while moving down would be conserved:

E=PE+KE = const.

where

PE is the potential energy

KE is the kinetic energy

However, this is not true if friction acts on the car. When the car is still at the top of the ramp, its speed is zero, so its kinetic energy is zero, and all the energy is just potential energy:

E=PE

While the car moves down along the ramp, friction does work on it, and part of the total mechanical energy is wasted and converted into thermal energy. As a result, the car reaches the bottom of the ramp with a final energy which is less than the initial energy:

E'

Also, at the bottom of the ramp, all the energy is just kinetic energy, since the height of the car is now zero, so

E'=KE

It follows that

KE

So, the potential energy of the car at the top of the ramp is greater than the kinetic energy of the car at the bottom of the ramp.

Learn more about potential and kinetic energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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I think it should be 'They are balanced' if I'm not wrong
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3 years ago
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What is the wavelength of the wave
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Explanation:

it is equal to the speed (v) of a wave train in a medium divided by its frequency (f): λ = v/f. Waves of different wavelengths.

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Read 2 more answers
A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

3 0
3 years ago
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X
Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

u'=0.625 c

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0
3 years ago
A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?
STatiana [176]
Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
P = 833.33 Watts

So, your final answer is 833.33 Watts
5 0
3 years ago
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