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polet [3.4K]
4 years ago
10

How long does it take to walk 0.13 miles

Physics
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

3 0
3 years ago
Segmented mirrors sag under their own weight. their optical shape must be controlled by computer-driven thrusters under the mirr
madam [21]
Active Optics.


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6 0
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
8 0
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:

The law of conservation of energy states that the total amount of energy in a system remains constant.

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
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