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Luda [366]
3 years ago
9

Kinetic energy can be passed from one object to another when objects collide,

Physics
2 answers:
antiseptic1488 [7]3 years ago
5 0
Kinetic energy can be passed from one object to another when objects collide,
Answer: True

Hope This Helps! :3
Dafna11 [192]3 years ago
3 0
True
`~~~~~~~~~~~~~~`
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1d. Conservation of energy is demonstrated in this roller coaster example.
Dvinal [7]

Answer:

can you clear your question I can't understand

6 0
3 years ago
A particle moves along the x axis from the origin. The magnitude of the position vector at time t is
Maurinko [17]

1) The average velocity is -2.1\cdot 10^5 m/s

2) The instantaneous velocity is 64t-260t^3

Explanation:

1)

The average velocity of an object is given by

v=\frac{d}{t}

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

x(t) = 32t^2 - 65t^4

where t is the time.

The position of the particle at time t = 6 sec is

x(6) = 32(6)^2 - 65(6)^4=-83,088 m

While the position at time t = 12 sec is

x(12)=32(12)^2-65(12)^4=-1,343,232 m

So, the displacement is

d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m

And therefore the average velocity is

v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

x(t) = 32t^2 - 65t^4

By differentiating with respect to t, we find the velocity vector:

v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

6 0
4 years ago
Two charges separated by one meter exert a 9 N force on each other. If the charges are pushed to a 3 meter separation, the force
tamaranim1 [39]

Answer:

False

Explanation:

The formula of force that exists between two charges is expressed as;

F = kq1q2/r²

If two charges separated by one meter exert a 9 N force on each other, the;

9 = kq1q2/1²

9 = kq1q2 ..... 1

If the charges are pushed to a 3 meter separation, then;

F =  kq1q2/3²

F =  kq1q2/9 .... 2

Divide both equations;

9/F = (kq1q2)/ kq1q2/9

9/F =  kq1q2 * 9/ kq1q2

9/F = 9

F = 9/9

F = 1N

Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False

3 0
3 years ago
A total of 875 J of work was done when a force of 175 N was exerted on a box to move it. How far was the box moved
Vladimir [108]
Work = Force times Distance
W = Fd

Given W = 750J, F = 125N;

750 = 125d

Solving for d:
d = 750/125
d = 6

The box moved a distance of 6 meters.
3 0
2 years ago
During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

7 0
3 years ago
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