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Levart [38]
3 years ago
6

A constant force of 5.00 N acts on a 2.50 kg object for 10.0 s. What are the changes in the object’s momentum and velocity?

Physics
1 answer:
dimulka [17.4K]3 years ago
7 0
Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct

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When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large
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Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

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. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa
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Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u>Vf = 128 ft/s</u>

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

<u>K.E = 122.8 Btu</u>

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