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Valentin [98]
2 years ago
15

If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn

er A has a
55.0-m head start?
Physics
1 answer:
yaroslaw [1]2 years ago
3 0

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A:    y = 7.50x + 55

Runner B:    y = 7.90 x

sooooo

  7.50 x + 55 = 7.90 x

- 7.50 x          - 7.50 x

 55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 =  1086.25 m

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A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
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Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

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3 years ago
A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force
Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

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applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

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therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

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