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Scilla [17]
3 years ago
13

Juan is standing on the street. An ambulance moves toward him and then passes by. What best describes the pitch that Juan hears?

The pitch remains constant as the ambulance moves toward and then away from Juan. The pitch increases and decreases in a repeated pattern as the ambulance passes Juan. The pitch drops to a lower pitch once the ambulance passes by Juan. The pitch increases to a higher pitch once the ambulance passes by Juan.
Physics
2 answers:
avanturin [10]3 years ago
7 0

Answer:

C) The pitch drops to a lower pitch once the ambulance passes by Juan.

Explanation:

ozzi3 years ago
5 0

Answer: The correct answer is "the pitch drops to a lower pitch once the ambulance passes by Juan".

Explanation:

Doppler effect is the phenomenon in which there is an increase or decrease in the frequency when there is relative motion between the source and listener.

Pitch depends on the frequency. if the frequency increases then pitch increases.

When the source and the listener are moving towards each other then there is increase in the frequency. When the source and the listener are moving away each other then there is decrease in the frequency.

In the given problem, Juan is standing on the street. An ambulance moves toward him and then passes by the pitch drops to a lower pitch once the ambulance passes by Juan.

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The acceleration due to gravity on the surface of Mars is gmars = 3.7 m/s2. How much would an 11 kg bag of potatoes weigh on Mar
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3 years ago
Define uniform and non uniform
cricket20 [7]

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7 0
2 years ago
Read 2 more answers
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find
snow_tiger [21]

Answer:

  • The emf of the generator is 6V
  • The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7   + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7   + 0.3r  

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

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