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3 years ago

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A particle of mass m moves along the x axis under the influence of a force given by F(x)=(3+2x)i. It starts from rest from the p

osition x=0. Find the magnitude of its acceleration and velocity at position x.

1 answer:

Aleks [24]3 years ago

5 0

Answer:

From Newton's second law of motion, acceleration can be calculated as follows:

F = ma

$F(x) = (3+2x)i = m a\\a_x=\frac{3+2x}{m}$

Use the third equation of motion to find the velocity at position x

$2as=v^2-u^2\\v=\sqrt{2a_xx+u^2}\\v=\sqrt{2\times \frac{3+2x}{m}\times x+0}\\v=\sqrt{\frac{6x+4x^2}{m}$

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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

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Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base

umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)Vf=17.08 m/s

Explanation:

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h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

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107=Vo*1.743

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3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

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Vf^2=292.04

Vf=17.08 m/s

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3 years ago