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Gnom [1K]
2 years ago
11

You are making a round trip from City A to City B and back to City A again at constant speed. At what point in the trip is your

average
speed equal to three times the magnitude of your average velocity?
Physics
1 answer:
MA_775_DIABLO [31]2 years ago
3 0

Answer:

Halfway between B and A on the return leg.

Explanation:

Your average SPEED for the entire trip will equal your constant speed as the time and distance increase at proportionate rates.

Your average VELOCITY will equal your constant speed while you travel from A to B because time and displacement are increasing at proportionate rates.

When you turn around at B to return, your Displacement is now decreasing while your travel time continues to increase, so your average velocity decreases.

Lets say the distance from A to B is 90 km and your constant speed is 30 km/hr.

your average speed is 30 km/hr because you took 6 hrs to travel 180 km

We want to find your position when your average velocity is 30/3 = 10 km/hr

it took 3 hrs to go 90 km from A to B. Let t be the time lapsed since turn around

your displacement is given by d = 90 - 30(t)

and your total time of travel is t + 3 hrs

 v = d/t

10 = (90 - 30t) / (t + 3)

10(t + 3) = (90 - 30t)

10t + 30 = 90 - 30t

40t = 60

t = 1.5 hrs

This will occur when you are halfway between B and A

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A rock has a mass of 25kg. The acceleration due to gravity is 10m/s and the height of the rock is 42 m. Find the potential energ
alisha [4.7K]

Answer:

 The potential energy of the rock = 10.5 kN

Explanation:

 Mass of rock = 25 kg

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You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out
kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0
3 years ago
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