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tino4ka555 [31]
3 years ago
5

An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard

by the observer if the source emits a 2550-Hz sound?
Physics
1 answer:
UNO [17]3 years ago
7 0

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,      

velocity of the observer = 17 m/s

speed of the sound = 343 m/s    

velocity of the source = 0 m/s    

frequency emitted from the source  = 2550 Hz              

f = f_0(\dfrac{v-v_0}{v-v_s})              

f = 2550\times (\dfrac{343+17}{343-0})

velocity of observer is negative as it is approaching the source.                   f = 2676.38 Hz ≈ 2677 Hz                    

hence, the frequency heard by the observer is equal to 2677 Hz

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2. Fill in the blanks
slamgirl [31]

Answer:

a. Two bodies are rubbed against each other they acquire equal and Opposite charge.

b. Electron are transferred from one object to another when two objects are charged by  friction.

c. Lightening rod is a device used to secure tall buildings from the effect of lightning

d. Negative charge is obtained by plastic comb when rubbed with dry hair.

e. Lightning is natural spark discharge on huge scale.

Hope this helps you. Do mark me as brainliest.

4 0
3 years ago
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
4 years ago
Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.
Mandarinka [93]

Answer:

4.9 x 10⁻⁴N

Explanation:

Given parameters:

Mass of the vertical strand of spiderweb  = 5 x 10⁻⁵kg

Unknown:

Tension in the vertical strand  = ?

Solution:

The tension is the vertical strand will be the weight of the strand.

  Weight of strand  = mg

m is the mass

g is the acceleration due to gravity = 9.8m/s²

Tension in the vertical strand  = 5 x 10⁻⁵kg  x 9.8m/s²

Tension in the vertical strand  = 4.9 x 10⁻⁴N

3 0
3 years ago
The product of an emf and a charge moving through it is:
liq [111]
The answer is to your question is force
3 0
3 years ago
While hammering a nail, you put in 30 Joules of work. The hammer then puts out 10 J of
AfilCa [17]

Answer:

C    Don't worry if you thought it was D. I almost answered that.

Explanation:

Remark

You can get this just by learning the vocabulary.

Efficiency = Work Out / Work In * 100%

Work in = 30J      That's you swinging the hammer.

Work out = 10 J    That's how much of the hammer's work gets to the nail.

Efficiency = (10/30)*100

Efficiency = 1/3 * 100

Efficiency = 33%

3 0
3 years ago
Read 2 more answers
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