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Greeley [361]
3 years ago
14

On hot sunny days, tire should not be over-inflated,why?

Physics
1 answer:
Alenkinab [10]3 years ago
4 0
The adds pressure to the tire which would cause it to burst
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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
Marianna [84]

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:

  • Distance form the sound source, s=50\ m
  • 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}

(C)

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}

(D)

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

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A calcium bromide compound is represented by which formula?
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Answer:

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Two objects feel 200 N of force due to gravity between them. If the mass of one object was doubled, what would be the new force
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Using a horizontal force of 60 N, a wagon is pushed horizontally across the floor a distance of 12 meters at a constant speed of
RideAnS [48]

Answer:

a) The work done by the force is 720 joules, b) The power supplied by the force is 138 watts.

Explanation:

a) Since force is uniform and parallel to the direction of motion, the work (W), in joules, done by the force (F), in newtons, is defined by this formula:

W = F\cdot s (1)

Where s is the travelled distance, in meters.

If we know that F = 60\,N and s = 12\,m, then the work done by the force is:

W = F\cdot s

W = (60\,N)\cdot (12\,m)

W = 720\,J

The work done by the force is 720 joules.

b) And an expression for the power supplied by the force (\dot W), in watts, is concieved by differentiating (1) in time:

\dot W = F\cdot \dot s

Where \dot s is the speed of the wagon, in meters per second.

If we know that F = 60\,N and \dot s = 2.3\,\frac{m}{s}, then the power supplied by the force is:

\dot W = F\cdot \dot s

\dot W = (60\,N)\cdot \left(2.3\,\frac{m}{s} \right)

\dot W = 138\,W

The power supplied by the force is 138 watts.

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