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2 years ago

15

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

its siren? The speed of sound on this day is 345 m/s.

1 answer:

Strike441 [17]2 years ago

8 0

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer $f_{obs}$ is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

$f_{obs} = f_s (\frac{v_w}{v_w-v_s})$

$f_s$ = Frequency of the source

$v_w$ = Speed of sound

$v_s$ = Speed of source

The velocity of the ambulance is

$v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$v_s = 30.55m/s$

Replacing at the expression to frequency of observer we have,

$f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})$

$f_{obs} = 878Hz$

Therefore the frequency receive by observer is 878Hz

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A person standing for a long time gets tired when he does not appear to do any work .why?

Keith_Richards [23]

Explanation:

A person standing still for a long time feels tired because the force of gravity acts on our body and puts stress on our muscles. so our muscles need energy to do work and keep body balanced and help to stand upright.

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2 years ago

A bicycle tire rotates 25 times in 10 seconds. What is it’s average angular velocity?

ryzh [129]

<h2>Answer:</h2>

<u>Angular velocity of bicycle tire is 15.78 radians per second.</u>

<h3>Explanation:</h3>

Angular velocity is the change in angular speed of an object with respect to time take for change or it is the rate of change of circular motion.

In the given question the circular displacement is 25 rounds around a central point.

The angular displacement is measured in degrees and 1 round is equal to 360 degrees.

25 Rounds = 25 × 360 = 9000 degrees.

Angular velocity = angular displacement /time = 9000/10 = 900 degrees per second.

In SI,angular velocity is represented in radians per second.

So, 1 radian = 57.29 degrees

Angular velocity = 15.78 radians per second

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2 years ago

a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

irinina [24]

Force , F = ma

F = m(v - u)/t

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s

F = 1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N

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1 year ago

You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the

steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units, 5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter / grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0

2 years ago

A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wh

otez555 [7]

Answer:

The percentage of the weight supported by the front wheel is A= 19.82 %

Explanation:

From the question we are told that

The center of gravity of the plane to its nose is $z = 2.58 m$

The distance of the front wheel of the plane to its nose is $l = 0.800\ m$

The distance of the main wheel of the plane to its nose is $e = 3.02 \ m$

At equilibrium the Torque about the nose of the airplane is mathematically represented as

$mg (z- l) - G_B *(e - l) = 0$

Where m is the mass of the airplane

$G_B$ is the weight of the airplane supported by the main wheel

So

$G_B =\frac{mg (z-l)}{(e - l)}$

Substituting values

$G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}$

$G_B = 0.8018 mg$

Now the weight supported at the frontal wheel is mathematically evaluated as

$G_F = mg - G_B$

Substituting values

$G_F = mg - 0.8018mg$

$G_F = (1 - 0.8018) mg$

$G_F = 0.1982 mg$

Now the weight of the airplane is = mg

Thus percentage of this weight supported by the front wheel is $A = 0. 1982 *100 =$ 19.82 %

7 0

2 years ago