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BlackZzzverrR [31]
2 years ago
8

If you are holding a book and you let go of it, why does it fall? Explain in terms of forces, potential energy, and kinetic ener

gy.
Physics
1 answer:
inessss [21]2 years ago
3 0

Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

Due to this process the whole energy is conserved.

As the potential energy decreases kinetic energy increases.

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Newtons. Force is mass times acceleration. Mass is measured in kilograms (kg) and acceleration is measured in meters per second squared (m/s^2.) These units (kg and m/s^2) multiplied together like in the equation force equals mass times acceleration (F=ma) gives a product with Newtons as the unit. 
6 0
3 years ago
You input 75 J of work with a wedge. If the wedge does 65 J of useful work, what if the efficiency of the wedge ?
zavuch27 [327]

Answer:

Efficiency of wedge is the ratio of "work done by the machine to the work supplied to the machine".

           Efficiency (η) = Work done by machine ÷ Work supplied

                                  = 65 ÷ 75

                                 = 0.86%

<em>Efficiency of wedge is 86%</em>

5 0
3 years ago
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of
Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

==> 1/3.742 = <em>0.27</em>

7 0
3 years ago
If the force acting on a cart doubles, what happens to the carts acceleration?
Effectus [21]

Answer:

If the force on a cart doubles, the acceleration of the cart doubles.

Explanation:

For this problem, we need to consider the following equation:

Force = Mass x Acceleration

We can reasonably assume that the cart will have constant mass in the given force system.  With this assumption we can say the following relationship:

Force is directly proportional to Acceleration within the system.

Given that our force on the cart is doubled, then our acceleration of the cart must also be doubled.  You can mathematically express this as follows:

F = MA

2F = M * 2A

Hence, if force doubles, the acceleration doubles.

Cheers.

8 0
3 years ago
To avoid using the variable "a"
sdas [7]

Answer:

option c .

Explanation:

..............................

6 0
2 years ago
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