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Assoli18 [71]
3 years ago
5

If the change in kinetic energy of a tennis ball hit by the racket

Physics
1 answer:
Marrrta [24]3 years ago
5 0

Answer: .36 m

Explanation:

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A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i
evablogger [386]

Answer:

The period of motion  of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

k = \frac{F}{x}

k = \frac{0.08829}{0.035}

k = 2.522 \frac{N}{m}

New mass(m_1) = 26 gm = 0.026 kg

Now the period of motion is given by

T = 2 \pi \sqrt{\frac{m}{k} }

T = 2 \pi \sqrt{\frac{0.026}{2.522} }

T = 0.637 sec

This is the period of motion  of new mass.

3 0
3 years ago
A 7 kg sled is initially at rest on a horizontal road. the sled is pulled a distance of 2.9 m by a force of 42 n applied to the
Artemon [7]
I have no idea on this question.
6 0
3 years ago
The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which da
RUDIKE [14]

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

5 0
3 years ago
Read 2 more answers
How many electrons are in the third energy level?<br><br> 2<br> 8<br> 18<br> 20
Blababa [14]

Answer:

B)8

Explanation:

In the first energy level you can have, at most, 2. Every energy level after that wants to have 8 electrons. Valence electrons I believe.

Hope this helps, have a nice day! (^-^)

6 0
3 years ago
Read 2 more answers
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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