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Zinaida [17]
3 years ago
12

Soccer ball is heading east positive direction with momentum of 2kgm/s. It is kicked backwards and a fraction of a second later

is moving west negative direction withmomentum of -2.3. What impulse did the kicker put on the ball on kgm/s?
Physics
1 answer:
attashe74 [19]3 years ago
8 0

Answer:

-4.3 kgm/s

Explanation:

Impulse which the soccer ball experiences is equal to the change in momentum of a body.

It is given mathematically as:

Impulse = m*Δv = mv - mu

Initial momentum of soccer ball = 2 kgm/s

Final momentum of the soccer ball = -2.3 kgm/s

Therefore, Impulse will be:

Impulse = -2.3 - 2

Impulse = -4.3 kgm/s

The impulse put on the soccer ball by the kicker is -4.3 kgm/s.

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<u>Explanation:</u>

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                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

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