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Marta_Voda [28]
3 years ago
9

A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine t

he acceleration of the object.
Physics
1 answer:
tatiyna3 years ago
8 0

Answer:

Explanation:

We will use the equation F - f = ma, which is just a fancy way of stating Newton's 2nd Law. For us:

F = 5.20 to the right (+)

f = 3.29 to the left (-)

m = 1.05 kg. Therefore,

5.20 - 3.29 = 1.05a and

1.91 = 1.05a so

a = 1.82 m/s/s to the right

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Answer:

The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

Explanation:

Given;

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Aleks04 [339]
The formula for the period of wave is: wave period is equals to 1 over the frequency.waveperiod=\frac{1}{frequency}
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

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get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s. 
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