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2 years ago

How can you produce more power than an excavator?

Physics

2 answers:

alina1380 [7]2 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]2 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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Three Small Identical Balls Have Charges -3 Times 10^-12 C, 8 Times 10^12 C And 4 Times 10^-12 C Respectively. They Are Brought

IgorLugansk [536]

Answer:

The charge in each ball will be 3 * 10^-12 C

Explanation:

(Assuming the correct charge of the second ball is 8 * 10^-12)

When the balls are brought in contact, all the charges are split evenly among then.

So first we need to find the total charge combined:

(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C

Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:

(9 * 10^-12) / 3 = 3 * 10^-12 C

So the charge in each ball will be 3 * 10^-12 C

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2 years ago

A car, X, with a mass of 1500 kg is moving west with an acceleration of 0.50 m/s2. A second car, Y, of equal mass is heading eas

ehidna [41]

Answer:

Explanation:

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2 years ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

Given:

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

Assume:

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

2 years ago

The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C

Mashcka [7]

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

-275*9.81=−2,697.75. Therefore, force A pulls downwards

Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

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2 years ago

How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?

slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

applied force, F = 600 N
mass of the truck, m = 1,500 kg
speed of the truck, v = 2 m/s

The force applied to the truck is determined by Newton's second law of motion; which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

$F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s$

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

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2 years ago