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andre [41]
3 years ago
10

An elephant pushes with 200 N on a load of trees. it then pushes these trees for 10 N. How much work did the elephant do?

Physics
1 answer:
Andreas93 [3]3 years ago
7 0

Answer:

the answer is 2000Nm

Explanation:

wprk done = force × distance moved

w.d = 200N × 10m

w.d = 2000Nm

mark me as brainliest plyyzzz

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butalik [34]

Explanation:

Recrystallization: contact pressure causing grains to "fuse" together

Cementation : precipitation of bonding agents between grains

Compaction : increase in density due to weight of overburden

Lithification is the process by which sediments are converted into sedimentary rocks. During this process, recrystallication, compaction and cementation of mineral grains occur.

The process starts with the compaction of sediments. The over burden weight of new sediments in the basin adds to the one originally deposited. This compresses the sediment. The volume of reduced and the density increases.

Recrystallization follows suit as the contact pressure of grains makes them fuse together. It is more like reworking of sediments. In this process, cementing materials can precipitate and cause sediments to be more fused together.

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4 0
3 years ago
What factor causes atmospheric pressure
bagirrra123 [75]
Atmospheric pressure is caused by the weight of the atmosphere pushing down on itself and on the surface below it.
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7 0
3 years ago
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What is the equivalent resistance of a circuit that contains three 10.0  resistors connected in series to a 6.0 V battery
disa [49]
The equivalent resistance of n resistors in series is given by:
R_{eq} = R_1 + R_2 + ... + R_n
In our circuit, we have three resistors of 10.0 \Omega each, therefore the equivalent resistance of the circuit is
R_{eq} = R_1 + R_2 + R_3 = 10.0 \Omega + 10.0 \Omega + 10.0 \Omega =30.0 \Omega
4 0
3 years ago
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What type of energy can take up space
Arada [10]

Answer:

Yes energy does take up space.

Explanation:

Every form of energy has a defining characteristic; sound is the vibration of molecules, electricity is the movement of electrons, and mass is the thing that take up space.

8 0
3 years ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
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