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Veseljchak [2.6K]
3 years ago
12

A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.

1 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops
Physics
1 answer:
shutvik [7]3 years ago
5 0

Answer:

The motorcycle travelled 69.73 m during these 3.1 s.

Explanation:

In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:

a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2

The distance is given by the following formula:

S = (V_final^2 - V_initial^2)/(2*a)

S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m

The motorcycle travelled 69.73 m during these 3.1 s.

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jim halves the distance between himself and a sound source. What is the change of decibels of the sound he hears?
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Answer:

6.02dB

Explanation:

The sound level intensity can be calculated by using the formula:

I(dB)=10log_{10}(\frac{I}{I_o})

where Io is the threshold of human hearing, and  I is the intensity of the sound in at a certain distance. I is given by

I=\frac{P}{4\pi r^2}

where P is the power of the source and r is the distance to the source. By replacing I in the expression for I(dB) we obtain

I(dB)=10log(\frac{P}{4\pi r^2I_o})=10log(\frac{P}{\pi r^2I_o})-10log(4)

when jim halves the distance we have r'=1/2r:

I'(dB)=10log(\frac{P}{4\pi (\frac{1}{2}r)^2I_o})=10log(\frac{P}{\pi r^2I_o})

Hence, the change o decibels is:

I'-I=10log(4)=6.02dB

hope this helps!!

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Answer:

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