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love history [14]
3 years ago
10

A ball is thrown downward; and its speed is v just before it strikes the ground. After the collision with the earth, it heads up

ward with a speed v. Which one of the following statements concerning this situation is truea) Only the magnitude of momentum of the earth is changed by collision (the correct answer, an explanation of why this is true would be very helpful)
b) the momentum of the earth is not changed by the collision
c) only the momentum of the ball is changed by the collision
d) the momentum of the ball is not changed by the collision
e) the collision causes both the momentum and the kinetic energy of the ball to change
Physics
1 answer:
d1i1m1o1n [39]3 years ago
4 0

For this case it is necessary to consider the assessments made between collisions of 'immovable' objects. In this case the earth is the immovable object or at least, it is approached by its mass. Considering the case of an inflatable ball of mass m that travels at a speed v towards the ground, it hits the floor and bounces with speed -v (Negative). The earth does not move, however, the momentum of the ball has changed by 2mv since the speed went from positive to negative. Applying the conservation of the momentum we know that the change of the momentum on the fly would be given under the function

p = m\Delta V

Considering the direction of the velocities this expression can be rewritten as

p = m (v-(-v))

p = 2mv

Therefore the correct answer would be

C. Only the momentum of the ball is changed by the collision

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Find the reaction supports at Ta and TB as shown in the loaded beam.
koban [17]

ANSWER

\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}

EXPLANATION

First, we have to make a sketch of the direction of the moments of the forces about 12m from the left in the diagram:

The sum of upward forces must be equal to the sum of downward forces. This implies that:

\begin{gathered} T_A+T_B=20+10 \\ T_A+T_B=30N \end{gathered}

Also, the sum of clockwise moments must be equal to the counter-clockwise moments:

\begin{gathered} (20\cdot8)+(T_B\cdot4)=(T_A\cdot12) \\ 160+4T_B=12T_A \\ \Rightarrow12T_A-4T_B=160 \end{gathered}

From the first equation, make TA the subject of the formula:

T_A=30-T_B

Substitute that into the second equation:

\begin{gathered} 12(30-T_B)-4T_B=160 \\ 360-12T_B-4T_B=160 \\ -16T_B=160-360=-200 \\ T_B=\frac{-200}{-16} \\ T_B=12.5N \end{gathered}

Substitute that into the equation for TA:

\begin{gathered} T_A=30-12.5 \\ T_A=17.5N \end{gathered}

Therefore, the reaction supports at TA and TB are:

\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}

5 0
1 year ago
What is formula for finding period of a planet if its mass is found by sending a spacecraft
Yanka [14]
T = D/V, where D = distance, V = tangential velocity.
7 0
2 years ago
At 9:13AM a car is traveling at 35 miles per hour. Two minutes later, the car istraveling at 85 miles per hour. Prove that are s
andreyandreev [35.5K]

Answer:

There is at least one instant which instantaneous acceleration is equal to average acceleration. a = 1500\,\frac{km}{h^{2}}.

Explanation:

The average acceleration experimented by the car is:

\bar a = \frac{85\,\frac{km}{h} - 35\,\frac{km}{h} }{\frac{2}{60}\,h }

\bar a = 1500\,\frac{km}{h^{2}}

According to the Rolle's Theorem, there is at least one instant t so that instantaneous acceleration equal to average acceleration for the analyzed interval. That is to say:

v'(c) = \frac{v(\frac{2}{60} )-v(0)}{\frac{2}{60}-0}

If car is accelerating at constant rate, instantaneous acceleration coincides with average acceleration for all instant t. Then, instantaneous acceleration is:

a = 1500\,\frac{km}{h^{2}}

6 0
3 years ago
Priya is responsible for collecting canned food along three different streets for her school's annual Thanksgiving Food Drive. S
AveGali [126]
Good luck sorry don’t know the answer
8 0
3 years ago
How do you find distance from average velocity and time
AlekseyPX

Answer:

Calculate the total distance travelled by the object - its motion is represented by the velocity-time graph below.

Here, the distance travelled can be found by calculating the total area of the shaded sections below the line.

½ × base × height.

½ × 4 × 8 = 16 m 2

(10 – 4) × 8 = 48 m 2

Explanation:

7 0
2 years ago
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