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3 years ago

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find the following Distance, Time, and Velocity in the problem; a track runs a 100-m dash in 11 seconds. What is his average vel

ocity during the run? (please include explanation and formula thankyou)

1 answer:

kow [346]3 years ago

5 0

Answer:

9.09m/s

Explanation:

now velocity of a body is measure of the distance covered by a body in a particular time frame,so simply its the rate at which distance is covered,

thus mathematically V(velocity)=d(distance covered)/t(time it takes to cover the distance)

hence V=100m/11s

which is V=9.09m/s.

so this means the runner covers 9.09m in every second ,<em> </em><em>well</em><em> </em><em>not</em><em> </em><em>bad</em><em> </em><em>lol</em><em> </em><em>!</em>

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Answer:

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Explanation:

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telo118 [61]

A = 4\pi r^2

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Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the

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Answer:

the magnitude of the work done by the two blocks is the same.

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$W_a = F_a d$

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3 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

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Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

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$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

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$a_{1}=\dfrac{v}{t}$

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Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

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We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

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3 years ago

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