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RSB [31]
3 years ago
5

Samuel hits a ball, which is initially at rest, off of a tee. What is the velocity of the ball after it is hit, if Samuel applie

s 30 J of work to the 1.1 kg ball? It will be helpful to show your work on a separate sheet of paper.
Physics
1 answer:
aniked [119]3 years ago
8 0

Answer:

7.39 m/s

Explanation:

Applying

K.E = 1/2mv²..................... Equation 1

Where K.E = Kinetic Energy, m = mass of the ball, v = velocity of the ball.

Make v the subject of the equation

v = √(2K.E/m)................. Euqation 2

From the question,

Given: K.E = 30 J, m = 1.1kg

Substitute these values into equation 2

v = √(2×30/1.1)

v = √54.54

v = 7.39 m/s

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Can a physical change, change what a substance is
Elden [556K]
No the substance will remain the same substance as before.
7 0
2 years ago
Can you please answers these for me please today is the last day to turn in work and I need this to pass please I’m begging than
Ymorist [56]

Answer:

1.   <u>F = ma</u>  <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N

2.  <u>F = ma</u>  <em>F - 18Kg * 3m/s² = 54Kg * m/s² =</em> 54N

3.  <u>F = ma</u>  <em>F = 0.025Kg * 5m/s² =</em> 0.125N

4.  <u>F = ma</u>  <em>F = 50Kg * 4m/s² =</em> 200N

5.  <u>F = ma</u>  <em>F = 70Kg * 4m/s² =</em> 280N

6.  <u>F = ma</u>  <em>F = 9Kg * 9.8m/s² =</em> 88.2N

Explanation:

Hope this helps ! ^^

8 0
2 years ago
Hanging from a horizontal beam are nine simple pendulums of the following lengths:
LenaWriter [7]

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

L =\frac{g}{w^{2} }

where g=9.8m/s

         ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

L= \frac{9.8}{2^{2} }

L = 2.45m

when  ω= 4rad/sec

L=\frac{9.8}{4^{2} }

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

8 0
3 years ago
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
pishuonlain [190]

Answer:

E_{total}=4.82*10^6N/C

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

E_{total}=E_1+E_2

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

E=\sigma/(2\epsilon_o)

Then:

E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0
3 years ago
Read 2 more answers
(01.01 LC) Two boxes are 8 cm apart. Which of the following should Janet do to
Trava [24]
Place the boxes 10cm apart, if they are closer that concentrates the mass of the boxes more
4 0
3 years ago
Read 2 more answers
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