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docker41 [41]
3 years ago
15

In each case, you should demonstrate how you worked out your answer, as well as giving the answer.

Physics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

First frequency f = 153.846 Hz

Second period T = 1.625 ms

Step by stepplanation:

T = 6.5ms= 6.5*(10^-3)

But T = 1/f

f = 1/T

f = 1(6.5*(10^-3))

f = 153.846 Hz

Now the frequency f is increased by a factor of 3

= 3*153.846

= 461.538

New frequency = 461.538+153.846

= 615.384 Hz

New period T = 1/615.384

Period T = 1.625*10^-3

T= 1.625 ms

irga5000 [103]3 years ago
3 0

Answer:

Explanation:

(a)

Time period, T1 = 6.5 ms

= 6.5 × 10^-3 s

Frequency is the reciprocal of time period.

f1 = 1 / T1 = 1 / (6.5 × 10^-3) = 153.85 Hz

Now the frequency becomes 3 times

f2 = 3 × f1 = 3 × 153.85 = 461.55 Hz

The new period is

T2 = 1 / f2 = 1 / 461.55

= 0.0022 second = 2.22 ms

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The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>
  • Initial velocity (u) = 8 m/s
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The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

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0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

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Learn more about motion under gravity:

brainly.com/question/13914606

5 0
2 years ago
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Nadusha1986 [10]
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8 0
3 years ago
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Answer:

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where

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Substituting numbers into the equation,

F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

The electrical force between the proton and the electron is given by

F_E=k\frac{q_p q_e}{r^2}

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In the case of the <u>Earth</u>, in which  <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy U_{p} will be:

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Where m is the mass of the object, g the acceleration due gravity and h the height of the object.

As we can see, the value of U_{p} is directly proportional to the height.

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The spring scale will read 559 Newton's or 125.7 pounds.
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