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3 years ago

14

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

's acceleration is 890 m/s2 and the vertical or y component of its acceleration is 1100 m/s2. The ball's mass is 0.41 kg. What is the magnitude of the net force acting on the soccer ball at this instant

1 answer:

mixas84 [53]3 years ago

4 0

Answer:

580.13N

Explanation:

The magnitude of the net force is expressed using the expression

F = ma

m is the mass

a is the net acceleration

Given

Mass = 0.41kg

Get the net acceleration

a = √(ax)²+(ay)²

a = √890²+1100²

a = √792100+1210000

a = √2002100

a = 1414.96m/s²

Get the required net force acting on the ball

F = ma

F = 0.41×1414.96

F = 580.13N

Hence the net force acting on the soccer ball at this instant is 580.13N

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A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

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Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

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sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

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Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

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