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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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Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len

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Answer:

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm$

Distance from second lens is 10 cm to the right

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm$

The final image will appear as +6.67 cm

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2 years ago

A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force

boyakko [2]

Answer:

Explanation:

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1 year ago

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The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i

Rainbow [258]

Answer:

$4.32\cdot 10^5 hp$ , $3.22\cdot 10^8 W$

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

$4F_T - F_d = 0$

where

$F_T = 322,000 N$ is the thrust force generated by each engine of the jet

$F_d$ is the drag force

Solving for Fd,

$F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N$

The velocity of the jet is

$v=250 m/s$

So, the rate at which the drag force does work (which is the power) is

$P=F_d v$

and substituting

$F_d = 1.288\cdot 10^6 N\\v = 250 m/s$

we find

$P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W$

Converting into horsepower,

$P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp$

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2 years ago

Write a hypothesis about the effect of the fan speed on the acceleration of the cart. Use the "if . . . then . . . because . . .

iragen [17]

As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
$s = ut + \frac{1}{2} at {}^{2}$
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time

Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
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2 years ago

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Answer:

$9.12\cdot 10^{-8} m$