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1 year ago

How can an object's motion be described and predicted?

1 answer:

3 0

Predicted motion is that motion in which the motion of the body or object is been predicted that the motion will be like this.

<h3>How can an object's motion be described and predicted?</h3>

According to the Newton's second law of motion say's that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. We can use it for the prediction of the motion of an object. The net force is the vector quantity which is the sum of all the forces which are acting upon the object.

So we can conclude that: Predicted motion is that motion in which the motion of the body or object is been predicted that the motion will be like this.

Learn more about Object Motion here: brainly.com/question/7074120

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You might be interested in

during a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative t

wel

The initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

<h3>What fraction of the initial mass of the spacecraft?</h3>

Increase the speed: Vf-Vi = 2.2 m/s

Speed of aircraft: Vr = 400 m/s

Speed of ejected products: Vrel = 1000 m/s

The answer is:

$V_f - V_i = V_{rel} log_{e} (\frac{mi}{mf})\\\\2.2 = 1000 log_{e} (\frac{mi}{mf})\\\\ log_{e} (\frac{mi}{mf} ) = \frac{2.2}{1000} \\\\ log_{e} (\frac{mi}{mf} ) = 0.0022$

$\frac{mi}{mf} = e^{0,0022} \\\\\frac{mf}{mi} = e^{-0,0022} \\\\\frac{mi-mf}{mi} = 1 - e^{-0,0022} = 0,00219$

So, the initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

Learn more about spaceship speed fraction brainly.com/question/28256735

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3 0

1 year ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago

A simple motor converts ________energy into________energy.

k0ka [10]

I think it is C. I hope I helped.

8 0

3 years ago

Read 2 more answers

Anyone know the answer ?

dlinn [17]

Momentum = mass x velocity, so 500kg x 2m/s = 1000 kg m/s

5 0

3 years ago

Read 2 more answers

A boat travels north across a river at a velocity of 22 meters/second with respect to the water. The river's velocity is 2.2 met

MArishka [77]

The magnitude of the resultant is

√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .

The direction of the resultant is

tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .

7 0

3 years ago