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jekas [21]
3 years ago
8

I need help in this pls i really need a answer

Physics
1 answer:
IrinaK [193]3 years ago
3 0

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\  = 30.0 + 10.0 + 20.0 \\  = 60.0 \: km

Total time is 3.0 hours

but:

average \: speed =  \frac{total \: distance}{total \: time}  \\

In terms of distance.

substitute:

average \: speed =  \frac{60.0}{3.0}  \\  \\  = 20 \:  {kmh}^{ - 1}

displacement = ( - 30.0) + 10 .0+ 20.0 \\  = 0

In terms of displacement:

speed =  \frac{0}{3}  \\  \\  = 0 \:  {kmh}^{ - 1}

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They weight the same

Explanation:

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Seasons are caused by differences in daylight, temperature, and weather patterns due to :
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... the angular tilt of the Earth's position on its axis relative to the sun


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What type of motion is shown with this graph? (5 points)
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A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
gavmur [86]
<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

5 0
3 years ago
Read 2 more answers
A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and
ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\

fo: frequency of the source = 440Hz

vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0
3 years ago
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