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defon
3 years ago
9

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.
Physics
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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Less than 7, Ph of 7 is neutral Acid Rain would be between 4 and 0. 
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What is mesothermal climates
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1 year ago
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:

Given:

initial temperature of water, T_i=18^{\circ}C

time taken to vapourize half a liter of water, t=18\ min=1080\ s

desity of water, \rho=1\ kg.L^{-1}

So, the givne mass of water, m=1\ kg

enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

where:

m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

Q_v=1.1282\times 10^{6}\ J

Now the power rating of the boiler:

P=\frac{Q_s+Q_v}{t}

P=\frac{342760+1128200}{1080}

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3 years ago
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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:

1)From second equation of motion we get

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

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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

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
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