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3 years ago

How can you produce more power than an excavator?

Physics

2 answers:

Tanya [424]3 years ago

8 0

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

sukhopar [10]3 years ago

8 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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A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla

Scorpion4ik [409]

Answer:

i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

2 L = n λ n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

v = λ f

λ = v / f

Let's replace in the first equation

2 L = 1 (v / f₁)

For the shortest length L = L-l

2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

v = 2L f₁

v = 2 (L-l) f₂

2L f₁ = 2 (L-l) f₂

L- l = L f₁ / f₂

l = L - L f₁ / f₂

l = L (1- f₁ / f₂)

Let's calculate

l / L = (1- 309/463)

i / L = 0.3326

4 0

3 years ago

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet

kogti [31]

The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.

4 0

3 years ago

Read 2 more answers

(a) what will an object weigh on the moon's surface if it weighs 100 n on earth's surface

juin [17]

We know the equation

weight = mass × gravity

To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.

So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)

Therefore,

100N = mass × 10
mass= 100N/10
mass= 10 kg

Now, all we need are the moon's gravitational field strength and to apply this to the equation

weight = 10kg × (gravity on moon)

4 0

3 years ago

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (

Crazy boy [7]

Answer:

Acceleration, $a=9.36\times 10^{18}\ m/s^2$