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tekilochka [14]
3 years ago
14

Hii please help i’ll give brainliest if you give a correct answer please

Physics
1 answer:
shusha [124]3 years ago
3 0

Answer:

D I think

Explanation:

Im just guessing dont trust 100%

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Suppose that an object travels from one point in space to another. Make a comparison between the magnitude of the displacement a
Rashid [163]

Answer:

- Distance is a scalar quantity, defined as the total amount of space covered by an object while moving between the final position and the initial position. Therefore, it depends on the path the object has taken: the distance will be minimum if the object has travelled in a straight line, while it will be larger if the object has taken a non-straight path.

- Displacement is a vector quantity, whose magnitude is equal to the distance (measured in a straight line) between the final position and the initial position of the object. Therefore, the displacement does NOT depend on the path taken, but only on the initial and final point of the motion.

If the object has travelled in a straight path, then the displacement is equal to the distance. In all other cases, the distance is always larger than the displacement.

A particular case is when an object travel in a circular motion. Assuming the object completes one full circle, we have:

- The distance is the circumference of the circle

- The displacement is zero, because the final point corresponds to the initial point

3 0
3 years ago
A 2-kg cart, traveling on a horizontal air track with a speed of 3m/s, collides with a stationary 4-kg cart. The carts stick tog
ladessa [460]

Answer:

The impulse exerted by one cart on the other has a magnitude of 4 N.s.

Explanation:

Given;

mass of the first cart, m₁ = 2 kg

initial speed of the first car, u₁ = 3 m/s

mass of the second cart, m₂ = 4 kg

initial speed of the second cart, u₂ = 0

Let the final speed of both carts = v, since they stick together after collision.

Apply the principle of conservation of momentum to determine v

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 3 + 0 = v(2 + 4)

6 = 6v

v = 1 m/s

Impulse is given by;

I = ft = mΔv = m(

The impulse exerted by the first cart on the second cart is given;

I = 2 (3 -1 )

I = 4 N.s

The impulse exerted by the second cart on the first cart is given;

I = 4(0-1)

I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).

Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.

8 0
4 years ago
How can force-time and force-displacement graphs be used to find the impulse or work done?
balandron [24]

Answer:

A. Area under force-time graph & Area under force-displacement graph

Explanation:

To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.

 Impulse  = Force x time

 Work done  = Force x displacement

When we plot a graph of force and time, the area under it is the impulse.

When a graph of force and displacement is plotted, the area under is the work done.

7 0
3 years ago
Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the c
ale4655 [162]

Answer:

TRUE

Explanation:

The answer is true.

Balance forces acting on a body will not change the motion of the body because the body experiences no net resultant force in one direction. When any body experiences equal forces with opposite directions, the net force or the resultant force experience by the body is zero.

In case of an unbalanced forces, there is a net force acting in one direction and so it causes the body to change in its state of motion in the direction of the net force.

4 0
3 years ago
A 0.75-kg ball falls vertically downward from a height of 55.0 m and rebounds upward. if the ball reaches a height of 30.0 m and
Nikitich [7]
Impuls I is given by:
I = \delta mv = |F|t
where \delta mv is the change in momentum, |F| is the average force and t is the time.

Solve the equation for the force F:
|F| = \frac{\delta mv}{t} = \frac{m(v_f - v_i)}{t}

Energy should be conserved, so the velocities will be:
\frac{1}{2}mv^2 = mgh \\ v = \sqrt{2gh}

Combining both equations:
|F| =  \frac{m( \sqrt{2g(h_f + h_i)} )}{t}
where h_f = 30m, h_i = 55m, m = 0.75kg, t = 0.0025s
4 0
4 years ago
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