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ra1l [238]
3 years ago
11

Starting from rest and moving in a straight line, a cheetah

Physics
1 answer:
UNO [17]3 years ago
3 0

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of  motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

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A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
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Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

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From law of conservation of energy,

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From Newton's equations of motion;

v² = u² + 2gh

Thus;

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(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

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The ejection seat has an acceleration of 8gees (8xgravity or ~80m/s/s). He has a mass of 70kg. The total force on him from the c
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To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the <em>net</em> force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)<em>g</em> ≈ 4900 N.

If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the <em>net</em> force would change to -5600 N - (70 kg)<em>g</em> ≈ -6300 N

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