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Vlad1618 [11]
2 years ago
10

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle

is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s .
Physics
1 answer:
vlada-n [284]2 years ago
6 0

frequency fapproach is heard by a passenger = 302.05 Hz

frequency frecede is heard by a passenger = 228.37433 Hz

Given :

speed of train to the ground = 30.0 m/s

frequency emitted by the train whistle = 262 Hz

speed of sound in air = 344 m/s

To Find :

(A) frequency fapproach (B) frequency frecede

Solution :

The frequency of approach is given by

f' = f <u>v + v(relative)</u>

v - v(air)

= 262x <u>344 + 18</u>

344 - 30

f' = 302.05 Hz

The frequency of approach is 302.05 Hz

The frequency of recede is given by

f' = f <u>v - v(relative)</u>

v + v(air)

= 262 x <u>344 - 18</u>

344 + 30

The frequency of recede is 228.37433 Hz

Learn more about Frequency here:

brainly.com/question/254161

#SPJ4

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3 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
Alexis is a scientist who is studying solid-state physics. Which activity would she most likely do as a part of this research? O
Ann [662]

Answer:

pretty sure its studying the atomic structure of a solid carbon dioxide. so c

Explanation:

8 0
3 years ago
Determine the kinetic of 600 kg roller coaster car that is moving with the speed of 35.2 m/s.
antiseptic1488 [7]
The answer is 10,560 Joules or 1.1*10^4


Explanation:

Step 1: Calculate
The equation for Kinetic Energy is

Kinetic energy=.5 times Mass times Velocity²
KE=.5*m*v²

so we plug in our numbers
KE=.5*600*35.2²

This works out to be 10,560 Joules or 1.1*10^4
5 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
4 years ago
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