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Nitella [24]
3 years ago
9

A cheetah runs at a constant velocity of 7 m/s. What is it’s acceleration in m/s/s PLEASE HELP

Physics
1 answer:
uysha [10]3 years ago
5 0

Answer:

0 m/s²

Explanation:

Acceleration is the change in velocity over change in time.  If the velocity is constant, then the acceleration is 0.

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a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i
Katyanochek1 [597]
Part A: 
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
             a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
            a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>

Part B: 
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
               F = m x a
where F is force,
m is mass, and 
a is acceleration

Substituting,
              F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

<em>ANSWER: 240 N </em>

Part C: 
The distance that the sprinter travel is calculated through the equation,
         d = V₀t + 0.5at²

Substituting,
            d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
             d = 13.5 m

<em>ANSWER: d = 13.5 m</em>

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3 years ago
The part of Earth's rocky outer layer that makes up the land masses is the
tangare [24]
D

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Order the sounds made by these sources from highest to lowest pitch.
Trava [24]

Answer:

c

motorcycle, telephone, piano, lawn mower

4 0
3 years ago
What is a material that reduces the flow of heat by conduction, convection, and radiation?
Oksi-84 [34.3K]
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4 0
4 years ago
Read 2 more answers
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
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