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inn [45]
2 years ago
15

A cheetah goes from 0 m/s to 25 m/s in 2.5s. what is the cheetahs rate of acceleration? A. 10m/s2

Physics
1 answer:
OleMash [197]2 years ago
5 0

The cheetah's rate of acceleration is 10 m/s².Option A is correct.

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u= 0 m/s

Final velocity, v= 25 m/s

Time elapsed, t = 2.5 seconds.

To find ;

Acceleration, a

The acceleration when the change in velocity is observed by the formula as:

\rm a= \frac{v-u}{t}

Substitute the given values:

\rm a= \frac{25-0}{2.5} \\\\  a= 10 \ m/s^2

The cheetah's rate of acceleration is 10 m/s²

Hence option A is correct.

To learn more about acceleration, refer to the link brainly.com/question/2437624

#SPJ1.

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Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A} , where:

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m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3

We divide the first equation by the second equation to get:

\frac{m}{R} = \frac{d A^2}{\rho}

A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8}   \ m^2

Using this Area, we find the diameter of the wire:

D = \sqrt{\frac{4A}{\pi}}

D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}

D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm

To find the length, we multiply the two equations stated initially:

mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}

l^2 = 8.534\\l =   2.92 \ m

8 0
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Answer: 90 kgm/s

Explanation:

The momentum (linear momentum) p is given by the following equation:

p=m.V

Where:

m=45 kg is the mass of the skater

V is the velocity

In this situation the skater has two values of momentum:

Initial momentum: p_{1}=m.V_{1}

Final momentum: p_{2}=m.V_{2}

Where:

V_{1}=3 m/s

V_{1}=5 m/s

So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

p=p_{2}-p_{1}=m.V_{2}-m.V_{1}

p=m(V_{2}-V_{1})

p=45 kg(5 m/s - 3 m/s)

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p=90 kgm/s

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