All categories

3 years ago

What are the 4 different types of mechanical waves

Physics

2 answers:

slamgirl [31]3 years ago

5 0

Answer:

Examples of mechanical waves

Water waves

sound waves

Spring waves

waves of the tuning fork

Explanation:

jolli1 [7]3 years ago

4 0

Answer:

Some of the most common examples of mechanical waves are water waves, sound waves, and seismic waves. There are three types of mechanical waves: transverse waves, longitudinal waves, and surface waves.

You might be interested in

If you and a friend are standing at opposites ends of a gymnasium and one of you claps, will the other hear the clap at the same

34kurt

Answer:

No. Your friend will not hear the clap when he/she sees it.

Explanation:

It takes time for sound waves to go through a large area. It takes longer if people are in the area rather than it is empty.

3 0

3 years ago

"a 1,600 kg car is traveling at a speed of 12.5 m/s. what is the kinetic energy of the car?"

xxTIMURxx [149]

I had this question before can you show me the answers choices

6 0

3 years ago

What is the result of convection currents in the atmosphere?

Degger [83]

Wind is the result of convection currents. :)

4 0

3 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. What is the dist

elena55 [62]

Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, $a=-44\ ft/s^2$

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}$

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

4 0

4 years ago

A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing

Damm [24]

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, $\tau _{shaft}$ = 30 MPa

Shearing stress of key, $\tau _{key}$ = 240 MPa

width of key, w = $\frac{d}{4}$

d is shaft diameter

Solution:

Torque, T = $\frac{P}{\omega }$

where,

$\omega = \frac{2\pi N}{60}$

$T = \frac{7460}{\frac{2\pi (1200 )}{60}}$ = 59.365 N-m

Now,

$\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}$

$30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}$

d = 0.0216 m

Now,

w = $\frac{d}{4}$ = $\frac{0.02116}{4}$ = 5.4 mm

Now, for shear stress in key

$\tau _{key}$ = $\frac{F}{wl}$

we know that

T = $F \times r$ = F. $\frac{d}{2}$

⇒ $\tau _{key}$ = $\frac{\frac{T}{\frac{d}{2}}}{wl}$

⇒ $240\times 10^{6}$ = $\frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}$

length of the rectangular key, l = 4.078 mm

7 0

3 years ago

Read 2 more answers