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sammy [17]
1 year ago
13

When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2

4 meters off the ground, its potential energy is
A.
the same.
B.
three times as much.
C.
twice as much.
D.
four times as much.
Physics
1 answer:
alexandr402 [8]1 year ago
8 0
I think the answer is twice as much
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ki77a [65]
<span>the answer would be 3,959 miles</span>
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A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)
castortr0y [4]

Answer: D.) 39,200 J

Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height.  In this case m = 400 kg, g = 9.8, h = 10 m thus:

P.E.=(400kg)(9.8\frac{m}{s^2} )(10m)=39,200 J

P.E.= 39,200 Joules

7 0
2 years ago
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The specific heat of aluminum is 0.90 J/gC . How much heat is given off when 25 grams of aluminum is cooled from 55 C to 25 C?
sp2606 [1]

Answer:

2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gºC? c=0.90J/g. 9 (2 sigs.)

Explanation:

8 0
2 years ago
A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho
Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>  

The ball rotates 6.78 revolutions.

     

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>        

At the bottom the ball has the following angular speed:

\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m

To find the revolutions we need the time, which can be found using the following equation:                

v_{f} = v_{0} + at  

t = \frac{v_{f} - v_{0}}{a} (1)

So first, we need to find the acceleration:

v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}    (2)  

By entering equation (2) into (1) we have:

t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}

Since it starts from rest (v₀ = 0):  

t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s

Finally, we can find the revolutions:  

\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

3 0
2 years ago
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

5 0
3 years ago
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