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lianna [129]
3 years ago
15

A net force of 6.8 N accelerates a 31 kg scooter across a level parking lot. What is the magnitude of the scooter’s acceleration

?a) 0.22 m/s2b) 0.69 m/s2c) 3.2 m/s2d) 4.6 m/s2
Physics
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

a)0.22 m/s².

Explanation:

Given that

Net force ,F= 6.8 N

mass ,m = 31 kg

From the second law of Newton's

F = m a   ---------------1

Where

F=Net force ,m=mass

a=Acceleration

Now by putting the values in the equation 1

F = m a

6.8 = 31 x a

a=\dfrac{6.8}{31}\ m/s^2

a = 0.219\ m/s^2

a = 0.22\ m/s^2

Therefore the acceleration of the scooter will be 0.22 m/s².

The answer will be "a".

a)0.22 m/s².

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
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Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

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Height = 1.94  m

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Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

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We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
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