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Vsevolod [243]
3 years ago
5

A parked car's horn (belonging to a musician) emits a concert A of frequency 440 on a day when the speed of sound is 342 m/s. Yo

u are in a car speeding along toward the parked car at a speed that makes the pitch appear to be A# (466HZ). At what speed in meters per second are you traveling?
Physics
1 answer:
julsineya [31]3 years ago
3 0

Answer:

19.08 m/s

Explanation:

f = actual frequency emitted by the parked car's horn = 440 Hz

V = speed of sound = 342 m/s

f' = frequency of the horn observed by you = 466 Hz

v = speed of your car moving towards the parked car = ?

frequency of the horn observed by you is given as

f' = \frac{Vf}{V - v}

466 = \frac{(342)(440)}{342 - v}

v = 19.08 m/s

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A load of 54 N attached to a spring that ishanging vertically stretches the spring 0.15 m.What is the spring constant?Answer in
beks73 [17]

Answer:

300 N/m

Explanation:

given,

Load attached to the spring, W = 54 N

length of stretch of the spring, x = 0.15 m

spring constant= ?

Force applied on the spring is calculated by the equation

F = k x

where k is the spring constant

x is the displacement of the spring due to applied load

now,

54 = k × 0.15

k = \dfrac{54}{0.15}

k =300\ N/m

hence, the spring constant is equal to 300 N/m

8 0
3 years ago
Read 2 more answers
A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
3 0
3 years ago
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A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closes
Marat540 [252]

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

3 0
2 years ago
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

7 0
3 years ago
Which statements about energy are true?
Masteriza [31]

Answer:

i think that the awnsre is d

7 0
3 years ago
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