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Anastaziya [24]
3 years ago
5

A stunt car traveling at 20 m/s flies horizontally off a cliff and lands 39.2 m from the base of the cliff. How tall is the clif

f?
Physics
1 answer:
Arada [10]3 years ago
7 0

Answer:

18.82 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Horizontal distance (s) = 39.2 m

Height (h) of the cliff =?

Next, we shall determine the time taken for the car to get to the ground. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Horizontal distance (s) = 39.2 m

Time (t) taken to reach the ground =?

s = ut

39.2 = 20 × t

Divide both side by 20

t = 39.2 / 20

t = 1.96 s

Finally, we shall determine the height of the cliff as follow:

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) taken to reach the ground = 1.96 s

Height (h) of the cliff =?

h = ½gt²

h = ½ × 9.8 × 1.96²

h = 4.9 × 3.8416

h = 18.82 m

Therefore, the cliff is 18.82 m high

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Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

Those who found this helpful please give me a Thanks to support me. So, I can explain other questions more clearly. If you don't want to mark me Brainliest don't mark. But, please give me a Thanks.

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Change in momentum is equal to force times the amount of time the force is applied? True or False?
harina [27]
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A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha
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Answer:

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Explanation:

Rotational kinetic energy is given by

\text{RKE} = \frac{1}{2}I\omega^2

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

I = \frac{2}{5}mr^2

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2

\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}

Translational kinetic energy is given by

\text{TKE} = \frac{1}{2} mv^2

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Read 2 more answers
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Are you starting to see the problem yet ?

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