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3 years ago

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2. Which are true about the relationship between momentum end energy?

1 answer:

lidiya [134]3 years ago

4 0

Answer:

Dependence on Velocity:

The momentum of an object is proportional to the object's velocity - if you double its velocity, you double its momentum. The kinetic energy of an object is proportional to the square of the object's velocity - if you double its velocity, you quadruple its kinetic energy.

Explanation:

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2. Because it can take one year to form just a few centimeters of new soil, 3 points

Grace [21]

Answer:

False.

Explanation:

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3 years ago

Three materials, put them in order of what

Grace [21]

Answer:

it's a

Explanation:

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3 years ago

A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much

nekit [7.7K]

Answer:

a $W_{3.5} = 490 \ J$

b $W_{2.5} = 250 \ J$

Explanation:

Generally the force constant is mathematically represented as

$k = \frac{F}{x}$

substituting values given in the question

=> $k = \frac{16}{0.2}$

=> $k = 80 \ N /m$

Generally the workdone in stretching the spring 3.5 m is mathematically represented as

$W_{3.5} = \frac{1}{2} * k * (3.5)^2$

=> $W_{3.5} = \frac{1}{2} * 80 * (3.5)^2$

=> $W_{3.5} = 490 \ J$

Generally the workdone in compressing the spring 2.5 m is mathematically represented as

$W_{2.5} = \frac{1}{2} * k * (2.5)^2$

=> $W_{2.5} = \frac{1}{2} * 80 * (2.5)^2$

=> $W_{2.5} = 250 \ J$

5 0

3 years ago

Water accelerated by a nozzle to 35 m/s strikes the vertical back surface of cart moving horizontally at a constant velocity of

tangare [24]

Answer:

a) The brake force = -535.7N

b) acceleration = - 1.2m/s^2

Explanation: Please find the attached files for the solution.

3 0

3 years ago

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

trapecia [35]

Answer:

a) $F_p=882N$

b) $P=4410W$

c) $V_p'=24135$ , $n=15.2\%$

Explanation:

From the question we are told that:

Mass $M=1500kg$

Velocity $v=4.9m/s$

Coefficient of Rolling Friction $\mu=0.06$

a)

Generally the equation for The Propulsion Force is mathematically given by

$F_p=\mu*mg$

$F_p=0.06*1500*9.81$

$F_p=882N$

b)

Therefore Power Required at

$V_p=5.0m/s$

$P=F_p*V_p$

$P=882*5$

$P=4410W$

c)

$V_p' =15mpg$

$V_p'=15*\frac{1609}$

$V_p'=24135$

Generally the equation for Work-done is mathematically given by

$W=F_p*V_p'$

$W=882*15*1609$

$W=2.13*10^7$

Therefore

Efficiency

$n=\frac{W}{E}*100\%$

Since

Energy in one gallon of gas is

$E=1.4*10^8J$

Therefore

$n=\frac{2.1*10^7}{1.4*10^8}*100\%$

$n=15.2\%$

7 0

3 years ago