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zvonat [6]
3 years ago
10

Determine explicitly which is faster, 75 miles per hour or 75 m/s? Express all your results in MKS units and explain your conclu

sions from the numbers, and explain using words. Show all work.
Physics
1 answer:
EleoNora [17]3 years ago
8 0

Answer:

75 m/s is faster

Explanation:

MKS units stands for meter kilogram seconds

75 miles per hour = 75 mph

1 mile = 1609.34 meters

1 hour = 60×60 = 3600 seconds

1 mph = 1609.34/3600 = 0.44704 m/s

75 mph = 75×0.44707 = 33.52792 m/s

Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.

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A power cycle operates between hot and cold reservoirs at 1200 K and 300 K, respectively. At steady state the cycle develops a p
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Answer:

Explanation:

a ) Thermal efficiency = work output / heat input

= .38 MW / 1 MW = .38

OR 38%

Heat rejected at cold reservoir = heat input - work output

1 MW - .38 MW

= 0.62 MW.

b ) For reversible power output

efficiency = T₂ - T₁ / T₂   ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.

= 1200 - 300 / 1200 = 900 / 1200

= .75

or 75%

rate at which heat is rejected

= 1 - .75 x 1

= .25 MW .

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3 years ago
Why does metal feel cold
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Explanation:

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2 years ago
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Jack pushed and pushed, and finally moved the new refrigerator into the kitchen. What is the BEST explanation of what happened?
Anika [276]

Answer:

a

Explanation:

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3 years ago
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3 years ago
To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar
Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

Explanation:

f = \frac{1}{2\pi*\sqrt{C*L}}

a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF  replacing:

f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}

f = 475.14 Hz

b. second smallest use the capacitive 6 uF so:

f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}

f = 1959Hz

c. second largest and largest oscillation first combination so:

Use 4.2 uF

f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{4.2uF*11mH}}

f = 2341.53 Hz

And finally largest oscillation cap in serie so:

C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}

C=2.47 uF

f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}

f =  3053.34 Hz

5 0
3 years ago
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