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Solnce55 [7]
3 years ago
14

How many students would it take to move 10 teachers, together exerting a force of 50 000N, once around a 400m track. Each studen

ts contributes 2000kJ of work.
Physics
1 answer:
PtichkaEL [24]3 years ago
4 0
Work done = force * distance that is move in direction of force

workdone = 50000N * 400m
= 2 * 10^7 J
since each student contributes 200kJ of work

no. of students = 2 * 10^7 J / 2000 * 10^3 J
= 10
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LenaWriter [7]

Answer:

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Explanation:

7 0
3 years ago
What is the energy equivalent of an object with a mass of 1. 05 g? 3. 15 Ă— 105 J 3. 15 Ă— 108 J 9. 45 Ă— 1013 J 9. 45 Ă— 1016 J
Molodets [167]

Considering the equivalence between mass and energy given by the expression of Einstein's theory of relativity, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

The equivalence between mass and energy is given by the expression of Einstein's theory of relativity, where the energy of a body at rest (E) is equal to its mass (m) multiplied by the speed of light (c) squared:

E=m×c²

This indicates that an increase or decrease in energy in a system correspondingly increases or decreases its mass, and an increase or decrease in mass corresponds to an increase or decrease in energy.  

In other words, a change in the amount of energy E, of an object is directly proportional to a change in its mass m.

In this case, you know:

  • m=1.05 kg
  • c= 3×10⁸ m/s

Replacing:

E= 1.05 kg× (3×10⁸ m/s)²

Solving:

<u><em>E= 9.45×10¹⁶ J</em></u>

Finally, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

Learn more:

  • brainly.com/question/9477556
5 0
2 years ago
The doppler effect is sensitive only to motion along the line of sight. <br> a. True <br> b. False
4vir4ik [10]
Its vey True trust me on this when I say it is
4 0
3 years ago
A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m
ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

S_{el} = 414Mpa

Yield Strain of the Specimen

\epsilon_{el} = 0.002

Diameter of the test-specimen

d_0 = 12.8mm

Gage length of the Specimen

L_0 = 50mm

Modulus of elasticity

E = \frac{S_{el}}{\epsilon_{el}}

E = \frac{414Mpa}{0.002}

E = 207Gpa

Strain energy per unit volume at the elastic limit is

U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}

U'_{el} = \frac{1}{2} (414)(0.002)

U'_{el} = 414kN\cdot m/m^3

Considering that the net strain energy of the sample is

U_{el} = U_{el}' \cdot (\text{Volume of sample})

U_{el} =  U_{el}'(\frac{\pi d_0^2}{4})(L_0)

U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})

U_{el} = 2.663N\cdot m

Therefore the net strain energy of the sample is 2.663N\codt m

6 0
3 years ago
When a force is applied to a wheel, its axle exerts a greater force?
makvit [3.9K]

Answer:

That is true.

Explanation:

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7 0
3 years ago
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