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anastassius [24]
3 years ago
5

A stationary siren creates an 894 Hz

Physics
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

12.3 m/s

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

Given:

fs = 894 Hz

fr = 926 Hz

c = 343 m/s

vs = 0 m/s

Find: vr

926 = 894 (343 + vr) / (343 + 0)

vr = 12.3

The speed of the car is 12.3 m/s.

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4 years ago
Help me please. Please see attached for the questions with the graph
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<u>Answer</u>

5) b-c

6)    a-b and

       e-f

7) f-g

9) a-b = 0 m/s

    c-d = 0.6667 m/s

    e-f = 0 m/s

    f-g = -3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

<u>Explanation</u>

Answer

5) b-c

In the section b-c the cart is accelerating because the slope of the graph is changing. The gradient that represent velocity is increasing.

6) a-b and e-f

At this sections the distance is not changing at all. This can only mean that the cart is not moving. It is at rest.

7) f-g

At this section the slope is negative meaning the cart is moving back to where it came from.

9) a-b = 0 m/s

At a-b the cart is not moving. So the velocity is zero.

<u>     c-d = 0.66667 m/s</u>

Velocity = distance / time

               =(50-40)/(40-25)

                = 10/15

                 = 0.6667  m/s

   <u> e-f = 0 m/s</u>

At e-f the cart is not moving. So the velocity is zero.

 <u>   f-g = -3 m/s</u>

Velocity = distance / time

               = (60-30)/(65-75)

                = 30/-10

                = - 3 m/s

10) b-c ⇒ The cart is acceleration.

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3 0
4 years ago
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

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