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2 years ago

What determines or dictates a inertia?

Physics

1 answer:

dolphi86 [110]2 years ago

4 0

<h2>Hope it <em>helps</em></h2><h2><em>have </em><em>a </em><em>nice</em><em> day</em><em> </em><em>;)</em></h2>

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A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

3 years ago

Which option most likely describes scientist(s) working in pure science?

lana [24]

Answer:

Two scientists in a lab examining vials of urine they are analyzing for levels of excreted protein.

Explanation:

3 0

2 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

3 years ago

If a child needs interventional service, he or she might have _____.

weqwewe [10]

Answer

Correct Answer is B,D, F and G

Explanation:

If a kid has any type of delays in his physical or mental growth then he need intervention service.

For example if a kid is not able to walk after 18 month, if a kid is not able to speak properly after 3 years, if a kid of 5 years age has lack of response to typical efforts to engage him in an interaction then these kids need intervention service.

8 0

3 years ago

FIND MEFIND V AT THE FIRST HILLFIND HEIGHT OF THE SECOND HILLFIND V AT POINT A

vitfil [10]

At the starting point, the spring releases potential energy which is converted to kinetic energy of the truck. The formula fr calculating the elastic potential energy of the spring is expressed as

PE = 1/2kx^2

where

x is the extension of the spring

k is the spring constant

From the information given,

k = 8500

x = 7

Thus,

PE = 1/2 x 8500 x 7^2 = 208250 J

Since elastic potential energy of spring = kinetic energy of the truck, it means that

Kinetic energy = 208250

The formula for calculating kinetic energy is expressed as

KE = 1/2mv^2

where

m = mass of truck

v = velocity of truck

From the diagram,

m = 600

Thus,

208250 = 1/2 x 600 x v^2

208250 = 300v^2

v^2 = 208250/300 = 694.17

v = square root of 694.17

v = 26.35 m/s

The velocity at which the truck is moving is 26.35 m/s

The potential energy of the truck at that point is calculated by apply the formula,

Potential energy = mgh

where

g = acceleration due to gravity and its value is 9.81 m/s

h is the height of the truck and it is 20

m is the mass of the truck and it is 600

Thus,

Potential energy = 600 x 9.81 x 20 = 117720

Mechanical energy = potential energy + kinetic energy

Mechanical energy = 208250 + 117720 = 325970 J

7 0

10 months ago