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nadya68 [22]
3 years ago
8

A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closes

t point. (a) Calculate the semimajor axis of the orbit. Incorrect: Your answer is incorrect. m (b) Calculate the eccentricity of the orbit. Incorrect: Your answer is incorrect. Did you find the semimajor axis a from the greatest and smallest radii
Physics
1 answer:
Marat540 [252]3 years ago
3 0

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

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8 0
3 years ago
Read 2 more answers
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
morpeh [17]

Answer:

v =4.36 m/s

Explanation:

given,

mass of object A = 18.5 Kg

initial velocity of object A = 8.15 m/s in east

mass of object B = 30.5 kg

initial velocity of object B = 5 m/s

P = P_A+P_B

P = m_Av_A\widehat{i} + m_B v_B\widehat{j}

P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}

P = 150.775 \widehat{i} + 152.5 \widehat{j}

P = \sqrt{150.775^2+152.5^2}

P = 214. 45 N s

velocity after collision is equal to

v =\dfrac{214.45}{18.5+30.5}

v =4.36 m/s

hence, velocity after collision is equal to 4.36 m/s

8 0
3 years ago
Read 2 more answers
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room t
emmasim [6.3K]

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

R_t = R_o[1 + \alpha(T-T_o)]\\\\

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T  = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

8 0
4 years ago
A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a friction
marissa [1.9K]

Answer:

Part a: <em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

Part b: <em>The total amount of energy transfer by the heat is 54.81 kJ</em>

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of CO_2 is given as

M=44 kg/kmol

the gas constant is given as

R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K

Volume of the tank is given as

V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3

Final mass is given as

m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg

Mass of the CO2 moved to the cylinder

m=m_1-m_3\\m=3-2=1 kg

The initial mass in the cylinder is given as

m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg

The mass after the process is

m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\

Now the volume 2 of the cylinder is given as

V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3

Part a:

So the Work done is given as

W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ

<em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

Part b:

The total energy transfer by heat is given as

Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ

As the temperature is constant thus change in internal energy is 0.

<em>The total amount of energy transfer by the heat is 54.81 kJ</em>

7 0
3 years ago
True false according to current atomic theory electrons are in fixed locations
Amanda [17]

That's <em>false.</em>  According to current atomic theory and quantum mechanics, an electron is always PROBABLY SOMEWHERE in a fuzzy foggy area around the nucleus of the atom, and that's all we can <u>ever</u> say about it.  There's no way to even TELL where the electron is, even if we <u>had</u> the technology to see things that small.

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3 years ago
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