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2 years ago

A 50-kg sprinter accelerates from 0 to 11 m/s in 3.0 s. What is the power output for this rapid start?

Physics

1 answer:

Citrus2011 [14]2 years ago

3 0

Answer:

1008.33 W.

Explanation:

Power: This can be defined as the rate at which energy is consumed or dissipated. The S.I unit of power is Watt (W).

P = E/t.......................... Equation 1

Where P = Energy, E = Energy, t = time.

But from the question,

E = 1/2mΔv²................... Equation 2

where m = mass of the sprinter, Δv = change in velocity of the sprinter = final velocity - initial velocity.

Substitute equation 2 into equation 1

P = 1/2mΔv/t....................... Equation 3

Given: m = 50 kg, Δv = 11-0 = 11 m/s, t = 3.0 s.

Substitute into equation 3

P = 1/2(50)(11)²/3

P = 25(121)/3

P = 1008.33 W.

Thus the power output = 1008.33 W.

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tekilochka [14]

This problem has three questions I believe:

> How hard does the floor push on the crate?

We have to find the net vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N

The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
= 1721.63 N

> Find the friction force on the crate

We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely, so we can say that the horizontal component is zero:

Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
= 811.66 N

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

μs = Fh / Fn

= 811.66 N / 1721.63 N

= 0.47

8 0

2 years ago

The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n

qaws [65]

Felectric = q*E
 Ftranslational = m*a
 Felectric = Ftranslational
 q*E = m*a
 Solve for a
 a = q/m*E 
 Our sign convention is "up is positive"
 q = 1.6*10^-19 C
 m = 1.67*10^-27 kg
 E = -150 N/C (- because it is down and up is positive)
 a = -6,4*10^5 m/s^2 (downward)
 answer
a = -6,4*10^5 m/s^2 (downward)

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2 years ago

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

BaLLatris [955]

The answer would be 2.8m height on earth takes
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2 years ago

Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i

fiasKO [112]

Answer:

The algebraic equation is:

$F_{v} =\frac{m_{b}v_{b}^{2} }{R} -m_{b} g$

Explanation:

Given information:

mb = book's mass

vb = tangential speed

R = radius of the path

Question: Derive an algebraic equation for the vertical force, Fv = ?

To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:

$F_{v} =\frac{m_{b}v_{b}^{2} }{R} -m_{b} g$

The variables were defined above and g is the gravity.