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3 years ago

Two football players each applied 100 n of force to sack the qb and move him 3 meters downfield. How much work was done?

1 answer:

4 0

<span>We know that the equation for work is W=force x distance, where work is in Joules, force is in Newtons and distance is in meters. In this situation, W= (100*2) x 3 which would result in W=600 J. The work done was 600 Joules.</span>

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Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle 0 between th

erik [133]

Based on the mass of the bob and the angle between the cord and the vertical, the pendulum length is 0.50m.

The maximum kinetic energy can be found to be 9.42 x 10⁻⁴J.

<h3>What is the pendulum length?</h3>

This can be found as:

= g-force / w²

Solving gives:

= 9.8 / 4.43²

= 0.4998 m

= 0.50 m

<h3>What is the maximum kinetic energy?</h3>

This can be found as:

= 0.5 × m × w² × A²

Maximum kinetic energy is:

= 0.5 × 60 × 10⁻³ × (4.43 × 0.4998 x 0.08 rad)²

= 9.42 x 10⁻⁴J

Find out more on maximum kinetic energy at brainly.com/question/24690095.

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2 years ago

I’ve already figured out A. I just need help with B and C.

Damm [24]

Answer:

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1456543.00

Explanation:

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3 years ago

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mrs_skeptik [129]

Then report it and it might be taken down

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3 years ago

Which scenario is in excample of a scientific way to think

vlabodo [156]

Can you explain this a bit more I don’t quite understand

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3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago