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kipiarov [429]
3 years ago
7

A loud sound is produced in the downtown section of a city. Which of the following is least likely to occur with the sound wave

s?
A. The sound wave will reflect off Buildings and automobiles.

B. The air will transmit the sound in longitudinal waves of energy.

C. All those sound waves will be absorbed by the surroundings.

D. The sound will bend spread between buildings by the fraction.
Physics
1 answer:
JulsSmile [24]3 years ago
6 0

Answer:

A. The sound wave will reflect off Buildings and automobiles.

Explanation:

This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.

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You observe a distant galaxy. You find that a spectral line, resulting from an electron transition in hydrogen, is shifted from
alexandr1967 [171]

Answer:

The galaxy is moving away from the observer

Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.

So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.

6 0
3 years ago
ID: A
postnew [5]

Answer: Velocity=8.26m/s

Explanation:

Acceleration=Finial velocity (V) - Initial velocity (u) ÷ Time

that is, a=v-u/t

a=1.2m/s², v=?, u=5.5m/s, t=2.3s

From a=v-u/t, make v the subject of the formula

v=at + u

v=(1.2* 2.3) + 5.5

v=2.76+5.5

v=8.26m/s

4 0
3 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

  • x0 is the initial position.
  • v0 is the initial velocity.
  • a is the acceleration.
  • t is the time interval in which the motion is studied.

In this case:

  • x0= 0
  • v0= 0  because the object is initially stationary
  • a= 6 \frac{m}{s^{2} }
  • t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 \frac{m}{s^{2} }*(15s)²

Solving:

x=½*6 \frac{m}{s^{2} }*(15s)²

x=½*6 \frac{m}{s^{2} }*225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0
3 years ago
A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Rou
Slav-nsk [51]

Answer:

3.6 seconds

Explanation:

Given:

y₀ = y = 0 m

v₀ = 31 sin 35° m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²

0 = 17.78t − 4.9t²

0 = t (17.78 − 4.9t)

t = 0 or 3.63

Rounded to the nearest tenth, the ball lands after 3.6 seconds.

4 0
3 years ago
Suppose you lift a 20 kg box by a height of 1.0 m.
9966 [12]
W=mgh W=(20)(9.8)(1) w=196J
6 0
3 years ago
Read 2 more answers
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