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Alex Ar [27]
3 years ago
12

A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el

astic, the ball bounces back with the same speed v going upward. Which of the following statements about the bounce are correct? (There could be more than one correct choice.) 1. The balrs mpmentum was conserved during the bounce. 2. The ball ha the same momentum just before and just after the bounce. 3. The magnitude of the ball's momentum was the same just before and just after the bounce. 4. All of the above 5. None of the above
Physics
1 answer:
artcher [175]3 years ago
7 0

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

                     p₀ = m v

The final moment

                     p_{f} = m (-v)

 

                  p₀ = - p_{f}

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
Which of the following experiments could be used to determine the inertial mass of a block? A. Place the block on a rough horizo
valentinak56 [21]

Answer:

D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force

Explanation:

For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.

3 0
3 years ago
I am pushing in a large box with 200 N of force. Clint is pushing on the box with 200 N of force in the opposite direction. Why
andreev551 [17]

Answer:

Balanced forces

Explanation:

Balanced forces are where two forces of equal size act on an object in opposite directions. It means that in each direction, any pushes and pulls are balanced by another force in the opposite direction.

4 0
3 years ago
Active immunity is passed on from a mother to her baby.
viva [34]
True.The immune system in babies. Antibodies are passed from mother to baby through the placenta during the last three months of pregnancy.
3 0
3 years ago
A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
USPshnik [31]

Answer:

Temperature increase = 2.1 [C]

Explanation:

We need to identify the initial data of the problem.

v = velocity of the copper sphere = 40 [m/s]

Cp = heat capacity = 387 [J/kg*C]

The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:

E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT  \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]

8 0
3 years ago
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