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Whitepunk [10]
3 years ago
11

Which of the following is an example of mechanical waves?

Physics
1 answer:
Nastasia [14]3 years ago
8 0
Sound Waves will be an example of mechanical waves.. hope this helps!
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A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr
Mazyrski [523]

Answer:

x = 0.0537 m or 5.37 cm

Explanation:

Given:

spring constant'k'= 4900 N/m

radius 'r' =0.029 m

Area 'A' =r²π = 0.029²π => 2.6 x 10^{-3} m²

Here, Pressure 'P' is given by,

Pressure = Force / Area

And we know that, for a spring :

F = kx, where k is the spring constant and x is the change in length.

P = kx/A

As P = 101325 Pa

101325 = 4900x / ( 2.6 x 10^{-3})

x = 0.0537 m or 5.37 cm

6 0
3 years ago
In 8.4 s a fisherman winds 2.9 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)
alukav5142 [94]

Answer:

The angular speed of the reel is 11.33 rad/s

Explanation:

Given

The fisherman takes t = 8.4 s to wind distance x = 2.9 m into a circle radius of r = 3 cm = 0.03 m

Than the tangencial speed equals the change in the distance to the time

v = \frac{x}{t} = \frac{2.9 m}{8.4 s} = 0.34 \frac{m}{s}

Knowing the tangencial velocity is proportional to the radius r and the angular velocity

v = r*w

w = \frac{0.34 m/s}{0.03 m} = 11.33 \frac{rad}{s}

3 0
3 years ago
A kind of foliated metamorphic rock a. coal b. limestone c. marble d. slate
djyliett [7]
Limestone is a kind of Foliated Metamorphic rock.  
4 0
3 years ago
Read 2 more answers
The orbital period of an object is 2 x 10^7 and its total radius is 4 x 10^4 m.
SVETLANKA909090 [29]

Answer:

12566

Explanation:

Use formula: vt=2pir/t

7 0
3 years ago
Read 2 more answers
State Pascal’s law. (2) b) The area of one end of a U-tube is 0.01 m2 and that of the other
Ray Of Light [21]

Answer:

Pascal Law's says that:

If the area of one end of a U-tube is A, and the area of the other end is A'. then if we apply a force F in the first end (the one of area A), the force experienced at the other end must be:

F' = F*(A'/A).

b) Now we can apply this to our particular case:

if the area of one end is 0.01m^2, and the area of the other end is 1m^2

Then we have:

A = 0.01m^2

A' = 1m^2

So, if now we apply a force F in the first end, the force experienced at the other end will be:

F' = F*(1m^2/0.01m^2) = F*100

This means that the force in the other end must be 100 times the force in the first end.

3 0
3 years ago
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