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earnstyle [38]
3 years ago
8

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from

its siren
Physics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer:

77.6 Hz

Explanation:

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren.

The given parameters are:

Observer speed = 115 km/h

Source frequency = 753 Hz

Speed of sound = 342 m/s

First convert km/h to m/s

Observer speed = (115 × 1000) / 3600

Observer speed = 31.94 m/s

The frequency received by the person will be:

F = fv / ( V - v )

Where

F = frequency received by the person

f = siren frequency

V = speed of sound

v = speed of the ambulance

Substitute all the parameters into the formula

F = (753 × 31.94) / ( 342 - 31.94 )

F = 24050.82 / 310.06

F = 77.568 Hz

Therefore, the frequency received by a person is approximately 77.6Hz

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anygoal [31]

True

Explanation:

A base is a substance that is often used as the chemical opposite of an acid.

Both behaves in opposite way to one another.

They can be said to complementary or conjugate chemicals.

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3 years ago
One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the flo
tekilochka [14]

Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x  .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

4 0
3 years ago
A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
3 years ago
What does it mean when it asks how many Macromolecules are in a diet?
Alona [7]
Do it as if you are writing a yes or no statement but have edge more towards the no side.
3 0
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Why does a lone pair of electrons occupy more space around a central atom than a bonding pair of electrons?
Inessa [10]

Answer:

The lone pair of electrons occupy more space because the electrostatic force becomes weaker.

Explanation:

When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.

In the case of the lone pair, only 1 nucleus is present, and the enticing electrostatic force becomes weaker and the intensity of the electrons will be increases. Therefore, the lone pair occupies more space than the pair of bonds.    

5 0
3 years ago
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