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agasfer [191]
3 years ago
6

A man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s2. What amount of force acted on the ball?

Physics
1 answer:
Delvig [45]3 years ago
4 0
Newton taught us that    Force = (mass) x (acceleration)

Force  =  (0.2) x (20) = <em>4 newtons</em> .

Something to think about:  The ball can only accelerate while the club-face
is in contact with it.  Once the ball leaves the club, it can't accelerate any more,
because the force against it is gone.
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You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

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The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

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Angular frequency of oscillation is given by:

\omega = 2\pi \times f

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\omega = 25.13 rad/s

5 0
4 years ago
Two identical speakers are emitting a constant tone that has a wavelength of 0.50 m. Speaker A is located to the left of speaker
Marianna [84]

Answer:

a) and c).

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d = |dA- dB| = (2n-1)*(λ/2)

For n= 1,  d = λ/2 = 0.25 m, it doesn't meet any of the cases.

For n=2, d= 3*(λ/2) = 0.75 m

In the case a) we have dA = 2.15 m and dB = 3.00 m, so dB-dA = 0.75 m, which means that in the location stated by case a) a complete destructive interference would occur.

For n=3, d= 5*(λ/2) = 5*0.25 m = 1.25 m.

This is just the case c) because we have dA = 3.75 m and dB = 2.50 m, so dA-dB = 1.25 m, which means that in the location stated by case c) a complete destructive interference would occur also.

The remaining cases don't meet the condition stated above, so the statements found to be true are a) and c),

5 0
3 years ago
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allsm [11]

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From the diagram

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