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svp [43]
3 years ago
15

Write down the effect of humidity and temperature on the speed of sound.​

Physics
1 answer:
finlep [7]3 years ago
4 0

Answer:

The speed of sound is affected by temperature and humidity. Because it is less dense, sound passes through hot air faster than it passes through cold air. ... The attenuation of sound in air is affected by the relative humidity. Dry air absorbs far more acoustical energy than does moist air.

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A wheelchair ramp is 5.2 m long and 0.8 m high. Calculate the ramp’s mechanical advantage
Arlecino [84]

Answer:

5

Explanation:

62

3 0
3 years ago
The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed
dmitriy555 [2]

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

Speed of light, v = 340 m/s

Frequency, f = ?

we know:

wave number = \frac{frequency}{speed of light}

20 = \frac{f}{340 m/s} \\\\f = 20 X 340 s^-^1\\\\f = 6800 Hz

Therefore, the frequency of the wave is 6800 Hz

4 0
2 years ago
Write down applications of mechanics​
harkovskaia [24]

Answer:

Explanation:

applied Mechanics and its Growing Utilisation of Theoretical Mechanics.\

Structural Engineering.

Hydraulics.  

Mechanical Engineering.  

External Fluid Dynamics.

Planetary Sciences.

Life Sciences.

7 0
2 years ago
A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent
solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}

The angular velocity is given as

w=at\\w=(0.905)(2.55)\\w=2.31rad/s

So the Kinetic Energy is given as

K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J

3 0
3 years ago
How do you do this problem?
kvasek [131]

Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

Δy = h

v = 0

a = -g

Find: v₀

v² = v₀² + 2aΔy

(0)² = v₀² + 2(-g)(h)

v₀ = √(2gh)

Now find the height reached if the projectile is launched at a 45° angle.

Given:

v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

v = 0

a = -g

Find: Δy

v² = v₀² + 2aΔy

(0)² = √(gh)² + 2(-g)Δy

2gΔy = gh

Δy = h/2

5 0
3 years ago
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