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Andrew [12]
2 years ago
13

Average speed equals distance divided by time. on a journey to the

Physics
1 answer:
Tanzania [10]2 years ago
6 0

The average speed will be 2.38×10⁶ m/sec.The average speed of an object indicates the pace at which it will traverse a distance. The metric unit of speed is the meter per second.

<h3>What is the average speed?</h3>

The total distance traveled by an object divided by the total time taken is the average speed.

The speed calculated at any particular instant of time is known as the instantaneous speed.

Given data;

Distance travelled = 4.12x10¹⁶ meter

Time period= 1.73x10¹⁰ sec

The average speed is found as

\rm V_{avg}= \frac{d}{t} \\\\\ \rm V_{avg}= \frac{4.12 \times 10^{16}}{1.73 \times 10^{10} } \\\\\ V_{avg}=2.38 \times 10^6 \ m/sec

Hence, the average speed will be 2.38×10⁶ m/sec.

To learn more about the average speed, refer to the link;

brainly.com/question/12322912

#SPJ1

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The wavelength of the wave is given by the letter A. And the amplitude of the wave is given by the letter B.
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3 years ago
A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
Which one will it be
liq [111]

Answer:

none

Explanation:

it's to high up to be affected by the gravity

3 0
3 years ago
Read 2 more answers
A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, reaches the ground in 12s. Cal
Ksju [112]
Use kinematic equations to solve:

1) yf = yo + vo*t + 1/2at²

yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time

yf - yo = vo*t + 1/2at²

yf - yo = h

vo = 0

Thus,

h = 1/2at²

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2) vf = vo + at

vo = 0

Thus,

vf = at

vf = (9.8)(12) = 117.6 m/s
4 0
4 years ago
A 5.50 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
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Answer:

Explanation:

position

y(t) = 2.80t + 0.61t³

velocity is the derivative of position

v(t) = 2.80 + 1.83t²

acceleration is the derivative of velocity

a(t) = 3.66t

F = ma = 5.50(3.66(4.10)) = 82.533 N

which should be rounded to no more than three significant digits and arguably only two due to the 0.61 factor.

F = 82.5 N or 83 N

Yes the units are Newtons, cannot tell what your system will accept. May not want the units at all.

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3 years ago
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