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SVETLANKA909090 [29]

2 years ago

14

Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.5 m/s. Assume

the acceleration is constant until the top speed is reached and is zero thereafter. Let the +- direction point in the direction the cheetah runs.
Express the cheetah's top speed top in miles per hour (mi/h).
top=
mi/h
Starting from a crouched position, how much time accel does it take a cheetah to reach its top speed, and what distance does it travel in that time?

accel=
s
=
m
If a cheetah sees a rabbit 131.0 m away, how much time total will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?

total=

1 answer:

insens350 [35]2 years ago

4 0

Answer:

boy if you dont get t f outa here

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a person was using a spanner to loosen a tight nut.Their mass was 50kg the spanner was 0.5m long i) what is the persons weight?

sineoko [7]

1) Weight of the person: 490 N

2) Maximum torque: 245 Nm

Explanation:

1)

The weight of a body is equal to the gravitational force exerted on the body; it is given by the equation

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m is the mass of the body

g is the acceleration due to gravity

For the person in this problem,

m = 50 kg is the mass

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Therefore, the weight of the person is

$F=(50)(9.8)=490 N$

2)

The turning force (also called torque) exerted by a force rotating an object is given by

$\tau = Fd sin \theta$

where

F is the magnitude of the force

d is the length of the arm (the distance between the force and the pivotal point)

$\theta$ is the angle between the direction of the force and the arm

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F = 490 N is the force applied (the weight of the person)

d = 0.5 m is the arm (the length of the spanner)

The maximum torque is obtained when $\theta=90^{\circ}$ , therefore it is:

$\tau=(490)(0.5)(sin 90^{\circ})=245 N\cdot m$

Learn more about weight:

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$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

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$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

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Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

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