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gregori [183]
3 years ago
11

a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

big push to the finish ?
Physics
2 answers:
Rama09 [41]3 years ago
5 0
Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2
goblinko [34]3 years ago
4 0

Answer:

0.027m/s²

Explanation:

Acceleration (in its simpler form) is defined as the change in velocity divided by the interval of time in which occurred that change. Mathematically:

a=\frac{v_{2}-v_{1} }{t}

Where:

v_{1}: initial velocity

v_{2}: final velocity

t: time needed to that increase in velocity

Plugging the given values in the equation:

a=\frac{3.5m/s-31m/s}{15s}

a=\frac{0.4m/s}{15s}

<em>a= 0.027m/s²</em>

So, her acceleration during her big push was 0.027m/s²

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Answer:

Kinetic energy is energy in motion so therefor if you increase the velocity or in your case speed the kinetic energy also has to increase

Explanation:

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A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2
garik1379 [7]
U1= 30m/s
u2= 23m/s
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A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?
Gemiola [76]

Answer:

3,500,000 J​

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4 0
3 years ago
The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a
jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A V_a=382\ V

Potential at V_c=785\ V

when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1

Change in Kinetic Energy of particle moving under the Potential From C to B

q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2

Divide 1 and 2 we get

\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}

on solving we get

V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c

V_b=\frac{743}{3}=247.67\ V

                     

4 0
3 years ago
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