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3 years ago

The result of a wave generator traveling faster than the speed of a wave is:

Physics

1 answer:

e-lub [12.9K]3 years ago

7 0

Answer:

Shock wave.

Explanation:

A wave that will travel faster than the speed of sound. It will generate regions where the pressure is high, in the medium in which the wave travels.

As the wave travels faster than the sound wave, in that medium compressional waves pile up and will result in regions of high pressure. The source that generates the shock waves will be moving ahead of the waves that are produced by it.

Suppose an air craft travels faster than the speed of sound, it will generate shock waves.

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What is meant by an electrical current

Varvara68 [4.7K]

Answer:an electrical Current can be defined as the free flow of electrons through a circuit

Explanation:

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3 years ago

The Old Faithful geyser found in Yellowstone Park is an example of _____.

katen-ka-za [31]

Answer:

cone geyser

Explanation:

The Old Faithful geyser is the oldest discovered geyser in the Yellowstone national park. The eruptions of the geyser are particularly predictable. It is a cone type geyser.

Cone geysers generally have a spout through which the water ejects out. When super heated water in the tube then the water starts to boil and form bubbles of steam, after this process the eruption takes place.

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2 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

lana [24]

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

8 0

2 years ago

An apple from the top branch of the tree and an apple from the bottom branch of a tree fall at the same time. Which apple will h

Vedmedyk [2.9K]

B. The apple from the bottom will hit the ground earlier. This is because an increase in height causes an increase in the time that the object will fall, and therefore will affect the final velocity of the falling object. Moreover, the reduction in velocity due to friction from the air should also be considered.

3 0

2 years ago