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SCORPION-xisa [38]
2 years ago
4

a man stands still on a moving escalator and a woman walks past him in the same direction as the escalator. to a stationary obse

rver the man has a speed of 0.2 m/s and the woman has a speed of 0.5 m/s from the frame or reference of the man on the escalator how fast is the woman walking?
Physics
2 answers:
ICE Princess25 [194]2 years ago
7 0

Answer:

0.3 m/s

Explanation:

Apex :)

laiz [17]2 years ago
4 0
Let Ve =  speed of escalator
Let  Vm = speed of the man (Vm=0 because he is stationary)
Let Vw =  speed of woman

According to the stationary observer,
Because the man stands still on the escalator, his speed is Ve = 0.2 m/s.
The speed of the woman is Ve + Vw = 0.5 m/s

Therefore
0.2 + Vw = 0.5
Vw = 0.3 m/

Answer:
The woman is walking at 0.3 m/s.

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The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p
zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

Explanation:

Given that,

Average pressure P=1.01\times10^{5}\ Pa

Radius of earth R_{E}=6.38\times10^{6}\ m

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

P=\dfrac{F}{A}  

F=PA

F=P\times 4\pi\times R_{E}^2

Where, P = pressure

A = area

Put the value into the formula

F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2

F=516.6\times10^{17}\ N

Hence, The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

8 0
2 years ago
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Differences between Pressure and upthrust​
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Answer:

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4 0
2 years ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

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Given:

Distance = 50 yard = 45.72 meter

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To find:

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Formula used:

speed = \frac{distance}{time}

Solution:

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speed = \frac{distance}{time}

Thus,

Time = \frac{distance}{speed}

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Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

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