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velikii [3]
1 year ago
14

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. How far away is the reflecting object? (c

= 3.0 × 10^8 m/s)
Physics
1 answer:
Assoli18 [71]1 year ago
5 0

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time =  20 μs = 20 * 10^{-6} s

let distance to the reflecting surface be = x

total distance travelled by pulse will be  = 2x

speed = 3.0 × 10^{8} m/s

distance = speed * time

2x = 3.0 × 10^{8} * 20 * 10^{-6}

   x = 3000 m

The reflecting object would be 3000 m away

To learn more about echo here

brainly.com/question/14861578?referrer=searchResults

#SPJ4

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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
I have no idea what to do. Plz help!
aleksklad [387]
Work is (force) times (distance). For Amy, you know both of them, and you can easily multiply them to find the amount of work. For Joe, the distance is zero, which should tell you all you need to know.
7 0
3 years ago
A transverse wave on a string is described by the wave function
svp [43]

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

Read more on transverse wave here

brainly.com/question/2516098

#SPJ4

5 0
1 year ago
When you whisper you produce a 10-dB sound.
masha68 [24]
All you would do is for a, 10 times 2 is 20 so it would be 20-dB 
For b, 10 times 4 is 40 so it would be 40-dB
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4 0
3 years ago
You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.7
HACTEHA [7]

Answer:

5.62 m/s

Explanation:

Newton's law of motion can be used to determine the maximum speed of the elevator. In the question, we are given:

Force exerted by the elevator (R) = 1.7 times the weight of the passenger (m*g)

Thus: R = 1.7*m*g

Distance (s) = 2.3 m

Newton's second law of motion: R - m*g = m*a

1.7*m*g - m*g = m*a

a = 0.7*m*g/m = 0.7*g = 0.7*9.8 = 6.86 m/s²

To determine the maximum speed:

v^{2} _{f} = v^{2} _{i} + 2as= 0 + 2(6.86)(2.3) = 31.556

v_{f} = \sqrt{31.556} = 5.62 m/s

Therefore, the elevator maximum speed is equivalent to 5.62 m/s.

8 0
3 years ago
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