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creativ13 [48]
2 years ago
9

A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C

alculate the total energy needed to place the satellite in orbit from the earth, assuming g=9.8N/kg at the earth's surface.
​
Physics
1 answer:
lubasha [3.4K]2 years ago
3 0

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

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In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor
VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596 kg-m2

I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2

by perpendicular axis theorem

I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2

4 0
3 years ago
HELP WILL GIVE BRAINLIEST IF CORRECT
anastassius [24]
It’s B i literally jus learned this
6 0
3 years ago
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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
3 years ago
In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s
serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

a = \dfrac{1.2 \times 10^5}{86}

a =1395.35\ m/s^2

Using equation of motion

v² = u² + 2 a s

s =\dfrac{v^2}{2a}

s =\dfrac{52^2}{2\times 1395.35}

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

8 0
3 years ago
A weather balloon is floating at a constant height above the earth when it releases a pack of instruments. If the pack hits the
andrezito [222]

Answer:

c 275 m

Explanation:

Given parameters:

Final velocity  = 73.5m/s

Unknown:

Height of fall  = ?

Solution:

Since the body is falling from rest, U = 0 or initial velocity is 0m/s. Then we use one of the kinematics equation to solve this problem.

           V²   = U²  + 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

           73.5²   = 0²  +  (2 x 9.8 x h)

         5402.25 = 19.6h

                h = 275.6m

4 0
3 years ago
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