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2 years ago

12

Difference between on pitch and frequency

1 answer:

shepuryov [24]2 years ago

8 0

Answer:

A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. I hope I got it correct !!

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Bending of light as it travels through water droplets will result in what

miv72 [106K]

Refraction is the bending of light<span> </span>

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3 years ago

Can a physical quantity have different dimensions in different systems of units?

Mars2501 [29]

Answer: yes a quantity have different dimensions in different system of units . No,because in different system of units doesn't change the quantity but it only changes the numerical.

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2 years ago

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The hypotenuse of a right triangle is 29.0 centimeters. The length of one of its legs is 20.0 centimeters. What is the length of

Kisachek [45]

square root (29 squared - 20 squared)

Pythagoras' theorem

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2 years ago

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A 5kg bag falls a verticle height of 10m before hitting the ground.

g100num [7]

Answer:

$u = 7m {s}^{ - 1}$

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

$me(1) = me(2)$

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

$m \times g \times h = \frac{1}{2} \times m \times u {}^{2} \\$

we divide both sides of the equation with mass to make the math easier.

$9.8 \times 10 = \frac{1}{2} \times u {}^{2} \\ \frac{98}{2} = u {}^{2} \\ u { }^{2} = 49 \\ u = 7$

7 0

3 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = $F_1=5.6 lb$

Area of the piston = $A_1=0.6 inches^2$

Force applied on the brakes = $F_2$

Area of the brakes = $A_2=1.8 inches^2$

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

$\frac{F_1}{A_2}=\frac{F_2}{A_2}$

$F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb$

16.8 lb is the force on the brake pad of one wheel.

5 0

3 years ago