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nikklg [1K]
3 years ago
6

Physics part 1 I need help answering these

Physics
1 answer:
alex41 [277]3 years ago
6 0

Answer:

lhchccyohchocohchocoyxtidts8ts58d85dtitxicyi

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
3 years ago
What do you mean by resistance of conductor?state it’s unit.
dalvyx [7]

Answer:

its unit is Ohm

Explanation:

Resistance means material which resist the passing current  through it and the value of resistance says how much the material is resisting the current and it temperature dependent and the unit is Ohm.

7 0
3 years ago
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
3 years ago
Which statements about acceleration are true?
bonufazy [111]
It's definitely not B or C. There are things missing from A and D so we can't narrow it down any farther.
4 0
3 years ago
A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of
Roman55 [17]

Answer:

T=9.42Nm

Explanation:

From the question we are told that:

Radius r= 0.5 m

Current I= 3.0 A

Normal vector n=\frac{(2i - j +2k)}{3}

Magnetic field B= (2i-6j) T

Generally the equation for Area is mathematically given by

 A=\pi r^2

 A=3.1415 *0.5^2

 A=0.7853 m^2

Generally the equation for Torque is mathematically given by

 T=A(i'*B)

Where

 i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}

 X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]

 X\ component\ of\ i'*B=12

Therefore

 T=0.7853*12

 T=9.42Nm

7 0
3 years ago
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