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dimaraw [331]
3 years ago
7

Two cars are raised to the same elevation on service- station lifts. If one car is twice as massive as the other, how do their p

otential energies compare?
Physics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

The potential energy of the more massive one is twice that of the other.

Explanation:

Potential energy is given by

<em>PE</em> = <em>mgh</em>

where <em>m</em> = mass of body, <em>g</em> = acceleration of gravity and <em>h</em> = height or elevation.

For the less massive car, let the mass be m_1. Then its <em>PE</em> is

PE_1 = m_1gh

For the massive car, let the mass be m_2.  Its <em>PE</em> is

PE_2 = m_2gh

But m_2 =2m_1

\therefore PE_2 = 2m_1gh = 2(m_1gh) = 2PE_1

Hence, the potential energy of the more massive one is twice that of the other.

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An astronaut is on the moon. He drops a hammer from a height of 3.2metres and it takes 2.0 seconds to reach the lunar landscape.
Anvisha [2.4K]

Answer:

1/6 m/s^2      ( about 1/6th gravity of Earth ( 9.81 m/s^2)

Explanation:

Displacement =  yo  +  vo t  - 1/2 a t^2

      -  3.2          = 0     +  0     - 1/2 a(2.0)^2

      -     3.2       =                -2a

             a = 3.2 / 2 = 1.6 m/s^2

6 0
2 years ago
If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo
vovangra [49]

Answer:

13 N

Explanation:

The Net Force of an object should be the difference between the forces applied to the object if the object is not in equilibrium. This object is not in equilibrium so therefore by finding the difference between the forces, you will find your answer. 20 N - 7 N = 13 N.

6 0
3 years ago
Read 2 more answers
A 2290 kg car traveling to the west at 22.3 m/s slows down uniformly. How long would it take the car to come to a stop if the fo
Elina [12.6K]

Answer:

5.72 s

Explanation:

From Newton's law, F = ma

The East is +ve direction, Hence,

F = +8930 N

m = 2290 kg

a = ?

8930 = 2290 × a

a = 8930/2290 = 3.90 m/s²

So, we will find the time it takes the car to stop using the equations of motion

a = 3.90 m/s²

u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)

v = final velocity of the car = 0 m/s (since the car comes to rest)

t = time taken for the car to come to rest = ?

v = u + at

0 = - 22.3 + (3.90)(t)

3.9t = 22.3

t = 5.72 s

5 0
3 years ago
Read 2 more answers
Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral
Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit  such that

P=VI=V^2/R

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

Current2=\frac{V}{2R}

Power2=Voltage * Current2

Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2

TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}

6 0
2 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required:  vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
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