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kondor19780726 [428]
3 years ago
14

You are walking on a moving walkway in the airport. The length of the walkway is 59.1 m. If your velocity relative to the walkwa

y is 2.35 m/s, and the walkway moves with a velocity of 1.85 m/s, how long will it take you to reach the other end of the walkway
Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

14.1seconds approx

Explanation:

Given data

Distance= 59.1m

Your velocity= 2.35m/s

Walkway velocity= 1.85m/s

Total velocity= 2.35+1.85= 4.2m/s

We know that

Velocity= distance/time

time= distance/velocity

substitute

time= 59.1/4.2

time= 14.07

time=14.1seconds approx

Hence the time is 14.1seconds approx

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Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecule
Colt1911 [192]

Answer:

8563732.58906 Pa

3992793.23326 Pa

5708.00923 J

Explanation:

V = Volume

N = Number of molecules = 3\times 6.023\times 10^{23}

T = Temperature = 300 K

b = 7\times 10^{-29}\ m^3

k_ = Boltzmann constant = 1.38\times 10^{-23}\ J/K

P = Pressure

We have the equation

P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa

The pressure is 8563732.58906 Pa

For isothermal expansion

P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa

The pressure is 3992793.23326 Pa

Work done is given by

dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J

The work done is 5708.00923 J

7 0
2 years ago
A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return
tensa zangetsu [6.8K]

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

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tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

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