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AnnZ [28]
3 years ago
7

A textbook is dropped from a second floor window and falls to the ground. What happens to the acceleration as the book falls to

the ground
-the acceleration increases
-the acceleration decreases
-the acceleration remains constant *​
Physics
1 answer:
Sav [38]3 years ago
6 0

Answer:

The correct option is;

The acceleration remains constant

Explanation:

The acceleration is due to the force of gravitational attraction between the text book and the Earth

According to Newton's law of gravitation, there is an attractive force between all objects given by the following relation;

F = G×M₁×m₂/r²

Where;

G, M₁, m₂, and r are constant such that we have;

G×M₁/r² = Constant = The acceleraton due to gravity, g

F = g×m₂

So the acceleration of the textbook as it is being attracted by the force of gravity towards the ground (Earth) is remains constant.

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Vlad1618 [11]

Answer:

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Explanation:

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PLEASE HELP ASAP!!!! What displacement did the object undergo in the time interval between t= 2.0s and t=8.0s?
Katyanochek1 [597]
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The action force is the balloon pushing the air out. What is the magnitude of the reaction force of the air pushing on the ballo
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4 years ago
Read 2 more answers
Stairway must have uniform riser height and tread depth; variations in riser height or tread depth shall not be over _______ inc
alexandr1967 [171]

Answer:

\frac{3}{8} inches

Explanation:

the variations in riser height or tread depth should not be grater than \frac{3}{8} inches that is equal to 9.5 mm but the maximum riser height should be the  \frac{81}{4} inch  but variation in riser height should not exceed to \frac{3}{8} inches. The minimum riser height should be 7 inches which is equal to the 178 mm

5 0
4 years ago
3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0
3 years ago
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