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3 years ago

If there are 1.609 km in a mile, convert 135 miles/hour into meters per second. There are 1000 m in a kilometer.

Physics

2 answers:

Effectus [21]3 years ago

4 0

Answer:

97.1037936

Explanation:

lukranit [14]3 years ago

4 0

Answer:

60.3375 m/ per sec

Explanation:

first multiply 1609 (the amount of meters in a mile) by 135( the total number of miles). this brings you to conclude there are a total of 217,215 meters in 135 miles.

So 135 mph = 217,215 meters per hour.

From here I took a simple route so as to not need a calculator.

217,215 meters = 60mins. (divide both sides by 2)

108,607.5 = 30mins. (divide both sides by 2)

54,303.75 = 15mins. (divide both sides by 15)

3620.25 = 1min

3620.25 = 60secs. (divide both sides by 60)

60.3375 = 1sec

60.3375 meters per second = 135miles per hr.

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

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$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

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As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

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where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

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3 years ago

Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that

stepan [7]

Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm

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3 years ago