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givi [52]
2 years ago
10

4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going

when it hit the ground?
Physics
1 answer:
vaieri [72.5K]2 years ago
8 0

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

mgh = \frac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 15}

v=\sqrt{294}

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

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What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?
sineoko [7]

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = 9.8 m/s^2

Mass of the object = m

Force = mass × acceleration

510 N=m\times 9.8 m/s^2

m=52.04 kg\approx 52 kg

The mass of an object is 52 kg.

8 0
3 years ago
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A measure of randomness or disorder
vodka [1.7K]
It is simply called Entropy.

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3 years ago
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The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
3 years ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
How long have advances in astronomy been occurring? Only for the past few years
solniwko [45]

I'd have to say that the list of choices doesn't go far enough.

Advances in Astronomy have been occurring for at least the past two millennia (2000 years).  Maybe longer.

3 0
4 years ago
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