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3 years ago

How can you produce more power than an excavator?

Physics

2 answers:

alina1380 [7]3 years ago

5 0

Just do energy spent divided by time to get your answer. With this we can say a human might be able to!

Andre45 [30]3 years ago

4 0

Answer:

If the work done by us is greater than the excavator in the given time then we can produce more power than excavator.

Explanation:

Power is defined as work done per unit time. Let’s take an example:

Case I:

If we do 300J work in 2 seconds and excavator 500J work in the same time then:

Power produced by us = 300/2 = 150 watt

Power produced by excavator = 500/2 = 250 watt

In this case power produced by excavator is more than us.

Case II:

If we do 500J work in 2 seconds and excavator 300J work in the same time then:

Power produced by us = 500/2 = 250 watt

Power produced by excavator = 300/2 = 150 watt

In this case power produced by us is more than excavator.

Hence, If the work done by us is greater than the excavator in the given time then we can produce more power than excavator. But it is very difficult to believe in reality.

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Answer:

4.1 m

Explanation:

10 kW = 10000 W

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The power yielded by the wind turbine can be calculated using the following formula

$P = \frac{1}{2} \rho v^3 A C_p$

where $\rho = 1.2 kg/m^3$ is the air density, v = 8.89 m/s is the wind speed, A is the swept area and $C_p = 0.45$ is the efficiency

$10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45$

$10000 = 190A$

$A = 10000 / 190 = 52.7 m^2$

The swept area is a circle with radius r being the blade length

$\pi r^2 = A = 52.7$

$r^2 = 52.7 / \pi = 16.79$

$r = \sqrt{16.79} = 4.1 m$

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3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

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3 years ago

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you

soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

$|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s$

The interaction time to avoid that the water balloon breaks is 0.029s

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3 years ago

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3 years ago

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Answer:

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