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antiseptic1488 [7]
3 years ago
13

An apple weighs at 1N. the net force on the apple when it is in free fall is?

Physics
1 answer:
kozerog [31]3 years ago
3 0

The weight is the magnitude of the gravitational force between
the Earth and the apple.  There's no reason for the force to change
 just because the apple happens to be in motion.  If it weighs 1N on
the scale, then it weighs 1N while it's sailing over the fence, 1N on
the teacher's desk, and 1N while it's falling from the tree.

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An airplane is moving at 350 km/hr. If a bomb is
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Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

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y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

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Where:

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V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
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