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4 years ago

10

How long does it take to walk 0.13 miles

1 answer:

LekaFEV [45]4 years ago

5 0

<span>0.13 miles are equal to 209 meters or 228.8 yards and the human walking speed is about 3.1 miles per hour, with all those calculations we can say that every 0.05 miles would take an average person about 1 minute, and if we add 0.10 + 0.03 it would be about 4-5 minutes to travel that short distance.</span>

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Anakin Skywalker's pod racer has a mass of 450 kg. If the top speed of this racer is 947

andriy [413]

Answer:

15,569,653.3 Joules(J)

Explanation:

The equation used to find Kinetic Energy (KE) is

KE = $\frac{1}{2}$ m $v^{2}$

You have been given

m = 450kg

v = 947km/h

KE = ???

Firstly, we need to convert the km/h into m/s as this is the unit used in the KE equation

This can be done by dividing by 3.6

947km/h = 263.056m/s

Substitute you values into the equation

KE = $\frac{1}{2}$ m $v^{2}$

KE = $\frac{1}{2}$ * 450 * $263.056^{2}$

KE = 15,569,653.3 Joules(J)

Round your answer as appropriate

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3 years ago

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madam [21]

Active Optics.

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3 years ago

*An inductor is capable of dissipating 50W of heat energy when a current 0.8A flows through it at a certain frequency. Calculate

ale4655 [162]

I think that your answer would be D

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4 years ago

The fact that the total amount of energy in a system remains constant is a(n) experiment. theory. hypothesis. Or law.

Murrr4er [49]

Answer:

<h3>law</h3>

Explanation:

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8 0

3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago