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3 years ago

What is formed when the rock making up a tectonic plate breaks and moves Select one of the options below as your answer:

Physics

1 answer:

dem82 [27]3 years ago

6 0

A. a fault hopes this helps you guys

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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Neporo4naja [7]

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2 m v₀²

v = √2 v₀

6 0

2 years ago

What is the acceleration of a 349 kg object that moved with a force of 750 N?

zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

where

f is the force

m is the mass

From the question we have

$a = \frac{750}{349} \\ = 2.14899713...$

We have the final answer as

Hope this helps you

4 0

2 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

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

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