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nignag [31]
2 years ago
11

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

Physics
1 answer:
Darina [25.2K]2 years ago
7 0

Answer:

<em>a. t = 2.02 s </em>

<em>b. d = 20.2 m</em>

Explanation:

<u>Horizontal Motion </u>

If an object is thrown horizontally from a height h with a speed v, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The time the object takes to hit the ground can be calculated as follows:

\displaystyle t=\sqrt{\frac{2h}{g}}

The time does not depend on the initial speed.

The range or maximum horizontal distance traveled by the object can be calculated by the equation:

\displaystyle d=v.t

The man standing on the edge of the h=20 m cliff throws a rock with an initial horizontal speed of v=10 m/s.

a.

The time taken by the rock to reach the ground is:

\displaystyle t=\sqrt{\frac{2*20}{9.8}}

\displaystyle t=\sqrt{4.0816}

t = 2.02 s

b.

The range is:

\displaystyle d=10\cdot 2.02

d = 20.2 m

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  • Initial velocity=u=0
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  • TIME TAKEN BY STONE TO HIT WATER=t
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Now

\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2

\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2

\\ \sf\longmapsto 500=5t^2

\\ \sf\longmapsto t^2=100

\\ \sf\longmapsto t=10s

Now

\\ \sf\longmapsto h=cT

\\ \sf\longmapsto T=\dfrac{h}{c}

\\ \sf\longmapsto T=\dfrac{500}{340}

\\ \sf\longmapsto T=1.47\approx 1.5s

Total time:-

\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s

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2 years ago
using newtons law a force of 250N is applied to an object that accelerates at a rate of 5M/s2 what is the mass of the object?
AURORKA [14]

Answer:

50 kg

Explanation:

F = ma

250 N = m (5 m/s²)

m = 50 kg

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2 years ago
When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do
Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

F=\frac{\Delta p}{\Delta t}

where

F is the force

\Delta p is the change in momentum

\Delta t=0.030 s is the time interval

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

v=281 km/h =78.1 m/s is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N

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2 years ago
Which type of force can be in the same it opposite direction?
LuckyWell [14K]

Answer:

C) unbalanced

Explanation:

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2 years ago
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Answer:

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Explanation:

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