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3 years ago

What is the kind of energy that comes from movement?

Physics

2 answers:

Sedbober [7]3 years ago

8 0

The movement is motion energy

Brainiest plz

Sergio [31]3 years ago

4 0

Answer:

motion energy

Explanation:

motion wnergy is the sum of potential and kinetic energy

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago

If the center of mass passes outside the area of support of an object, what will happen to it?

defon

Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.

6 0

3 years ago

What if the Earth was moved out to Jupiter's orbit which is about 5

Stels [109]

Answer:

25.0 less force

Explanation:

4 0

3 years ago

Suppose you are measuring double‑slit interference patterns using an optics kit that contains the following options that you can

svetlana [45]

Answer:

3.6 m

Explanation:

$\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \ 0.25 \ mm\\\\= 0.25 *10^{-3} m$

$\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\$

Also

$\beta = 1 \ mm \ fringe \ width$

$D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} = 3.6 m$

Therefore, the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m

4 0

3 years ago

Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer

tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0

3 years ago