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shtirl [24]
3 years ago
6

A ball of mass m and speed v travels horizontally, hits a wall, and rebounds with the same speed as it had before the collision.

Determine the magnitude of the ball's change in momentum as a result of the collision.
Physics
2 answers:
larisa [96]3 years ago
6 0

Answer:2mv

Explanation:

Given

Ball travelling with speed v hits the wall and rebound with same speed

Initial momentum of ball is P_i=m\cdot v

Momentum after the collision is P_f=m\cdot (-v )

Change in momentum is \Delta P=P_i-P_f

\Delta P=mv-(-mv)

\Delta P=2mv

Therefore magnitude of change in momentum is 2 mv  

Bogdan [553]3 years ago
5 0

Answer: (MbxUb) + MbVb)= M( V+U). Where U=0, initial velocity.

It can be express as M1V1=M2V2.

Meaning momentum before collision is equal moment after collision.

Explanation:

If the ball returns with the same speed, definitely there will be no loss in the kinetic energy, so it returns with the same momentum before the impact.

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Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.
4 0
3 years ago
Read 2 more answers
You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque
harina [27]

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

                 Fe - W = m a  

                 k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

                Fe y - W y = I α

angular and linear acceleration are related

       a = α r

       Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0
3 years ago
Kissa and Lily ran a race in gym class Lily was off balance so she did not start until 2 seconds had passed
Paul [167]
Average speed = (distance covered) / (time to cover the distance)

Tissa covered 60 meters in 10 seconds. Her average speed was

(60 m) / (10 sec) = 6 m/s.

That's the slope of the dotted line.

Lilly covered 60 meters in 8 seconds. Her average speed was

(60 m) / (8 sec) = 7.5 m/s .

That's the slope of the solid line.

Lilly covered the same distance in less time, and both girls
arrived at the finish line together. Technically, in science talk,
we would say that Lilly ran "faster", and her average speed
was "greater".

We can detect that by looking at the graph, because Lilly's line
has the characteristic of being "steeper", and we know that the
slope of the line on a distance/time graph is "speed".
3 0
3 years ago
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.
pickupchik [31]
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction. 
AMA (Actual mechanical advantage) is found by dividing output force by effort force. The actual mechanical advantage will always be less than the ideal mechanical advantage. The ideal mechanical advantage assumes perfect efficiency which doesn't account for friction, while actual mechanical advantage does. Therefore; the IMA is always greater than the actual mechanical advantage because all machines must overcome friction. 
5 0
3 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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