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saul85 [17]
3 years ago
13

Two cars are sitting still. Car A has a mass of 2000 kilograms. Car B has a mass of 4000 kg. If the goal is to push both cars to

roll at the same speed, describe the force it will take to get Car B to that speed? (Assume everything else is equal: friction & steepness of the road, time, person pushing, etc.)
Physics
1 answer:
astra-53 [7]3 years ago
7 0
Since car B has twice as much mass as car A has, it would require 2x as much force that it would need to push car A down.

If that was kinda confusing lemme say it again,

It would take twice as much that would be needed to push car B down than car A

Since everything is the same, basically two people would need to push down car B and one person for car A

Hope this helps
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True.

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Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
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Answer:

U=50.96J

Explanation:

The electrostatic potential energy for pair of charge is given by

U=1/4π∈₀×(q₁q₂/r)

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U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

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U=3\frac{1}{4\pi E_{o} }\frac{q^{2} }{r}\\  U=3\frac{1}{4\pi8.85*10^{-12} }\frac{(1.15*10^{-6}C )^{2} }{0.0007m}\\  U=50.96J

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· A hot, just-minted copper coin is placed in 101 g of water to cool. The water
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hope this help you

3 0
3 years ago
Read 2 more answers
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

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3 years ago
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