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coldgirl [10]
4 years ago
9

A car travels 40 miles in 30 min. Calculate the car's average speed in mi/hr.

Physics
2 answers:
Law Incorporation [45]4 years ago
8 0
Speed = distance by time
your time is 30 min or 0.5h
you speed is 40 miles divided by 0.5h equals 80 miles per hour 

answer is 80mi/hr
Ivenika [448]4 years ago
7 0
1.33 is the answer
40/30 is 1.33
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Explanation:

Velocity = displacement / time

v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)

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7 0
3 years ago
Which phrase best describes sir isaac newton's contributions to modern science and therefore to the industrial revolution?
astraxan [27]
Can you please give the phrases? 

But, I'll help what I can.

First, he was the first to discover gravity. He was not bonked by the head by an apple, rather he watched an apple fall from a tree before he decided to explore gravity further. 

He was also the first scientist to be knighted, which is a great honor, as you can expect. 

Newton also developed The Three Laws of Motion. They are extremely important to physics and are considered some of the foundation for physics today. 

He also discovered calculus, which is complex math that is very helpful to scientists today. 

He also discovered the color spectrum using a glass prism, a dark room and window shade with a hole in it. He was able to project the color spectrum onto a piece of paper.

Those are the few I can think of now, but hope it helps!


7 0
3 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
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Answer:

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Explanation:

While many people would like to simply add the forces from each end to get a total force, this is fundamentally incorrect.

MIGHT BE TOTALLY WRONG

4 0
3 years ago
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