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Lady bird [3.3K]

3 years ago

3

The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force act

ing along the direction of motion. First, let us consider a particle of mass m moving along the x axis. The net force F is acting on the particle along the x axis. F is a constant force. Part A The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Express your answer in terms of any or all of m, F, and t.

1 answer:

m_a_m_a [10]3 years ago

4 0

Answer:

$p=F\times t$

Explanation:

We are given:

mass of the particle moving along x-axis = m

force acting on paricle along x-axis = F

time = t

momentum, p = ?

<u>From the Newton's second law of motion:</u>

$F=\frac{dp}{dt}$

where:

dp & dt are the change in momentum and change in time respectively.

Using the above eq. of the Newton's second law of motion:

$dp=F\times dt$

according to given:

$p=F\times t$

We also have for impulse:

$i=F\times t$

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What is the potential energy at point e when the total mechanical energy is 4900 for the 100 kg moving at 2 m/ s

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Answer:

B. 4700 J

Explanation:

Given the following data;

Mechanical energy = 4900J

Mass = 100kg

Velocity = 2m/s

To find the potential energy;

Mechanical energy = kinetic energy + potential energy

First of all, we would determine the kinetic energy of the object;

K.E = ½mv²

K.E = ½*100*2²

K.E = 50*4

K.E = 200 J

Substituting into the equation, we have;

4900 = 200 + P.E

P.E = 4900 - 200

P.E = 4700 Joules

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Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.

nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is $\frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2$

<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

Translational kinetic energy. Due to the linear movement of the body

$K_{tras} =\frac{1}{2} m v^2$

Rotational kinetic energy. Due to the rotational movement of the body.

$K_{rot} = \frac{1}{2} I w^2$

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

$K_{earth} = K_{tras} + K_{rot}$

Linear and rotational speed are related.

v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

I = $\frac{2}{5 } m r^2$

Let's subatitute.

$K_{earth} = \frac{1}{2} \ m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}$

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

$w = \frac{2 \pi}{T}$

Let's substitute.

$K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2 \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2$

$K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)$

Data for Earth are tabulated:

Mass m = 5.98 1024 kg

Radius r = 6.37 10⁶ m

Radius orbits tras = 1.496 10¹¹ m

Rotation period $T_{rot}$ = 24 h ( $\frac{3600s}{1h}$ ) = 8.64 10⁴s

Translation period $T_{tras}$ = 365 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 3.15 10⁷ s

Let's calculate.

$K_{earth} = 4 \pi^2 5.98 \ 10^{24} ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2 \ + \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )$

$K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)$

$K_{earth}= 2.66 \ 10^{33} J$

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of $T_{Sum}$ = 27 days.

The kinetic energy of the sun is;

$K_{sum} = K_{rot} = \frac{1}{2} I w^2$

$K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2$

$K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2$

The tabulated data for the sun are:

Mass m = 1,991 1030 kg.

Radius R = 6.96 10⁸ m

Period T = 27 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 2.33 10⁶ s

Let's calculate.

$K_{sum} = 1.40 \ 10^{36} J$

The relationship of the kinetic energy of the sun and the Earth is:

$\frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}$

$\frac{K_{sum}}{K_{earth}} = 5.3 \ 10^2$

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is: $\frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2$

Learn more about kinetic energy here: brainly.com/question/25959744

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Answer:

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4 years ago

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