Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.
Answers
To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.
Mathematically the conservation of these two energies can be given through
Where,
W = Work
Final gravitational Potential energy
Initial gravitational Potential energy
When the spacecraft of mass m is on the surface of the earth then the energy possessed by it
Where
M = mass of earth
m = Mass of spacecraft
R = Radius of earth
Let the spacecraft is now in an orbit whose attitude is then the energy possessed by the spacecraft is
Work needed to put it in orbit is the difference between the above two
Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is
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Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1? B. What is the change in the total momentum of the pair? C. What is the magnitude of the change in the momentum Δp2, of mass M2?
Answers
Answer:
All are true except the last point that says that a battery is a current source and the current at the outlet is always the same.
Explanation:
- A battery is an electro-chemical device which converts the chemical energy into usable electrical energy thus it provides electrical energy.
- Since, the battery maintains a a constant potential difference between its terminals, once connected.
- Since, the movement of electric current requires energy, which is supplied by the electric potential energy stored in the battery.
- The current in the battery flows as per the Ohm's law and we can not say that the current leaving will always remain constant.
- As the current is the flow of electric charge, and charges are not stored in batteries unlike capacitors, thus the current at the leaving end will depend on Ohm's law and will vary accordingly.
Answers
Answer: The volume is decreasing at a rate of 80 cm3/min
Explanation: Please see the attachments below
Answer: 80 cm³/min
Explanation:
Just solved it
Answers
E = hv, where v = the frequency and h = planck's constant.
So E = 5.20 x10^14 x 6.63 x 10^-34
= 3.45 x 10^-19 Joules
°C = 5/9 * (°F - 32°)
1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:
1. 5 cm = ________ mm
2. 83 cm = ________ m
3. 459 L = _______ ml
4. .378 Kg = ______ g
5. 45°F = ________ °C
6. 80°C = _________ °F
Answers
Answers
Answer:
So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.
Explanation:
Fb = Fg
so
qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6
= 0.115C
note:2800ft/sec = 853.44m/s
So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.
Answers
Answer:
66.053m/s
Explanation:
A = 47
B = 347
C = 19
Train moves at
(23 + A)m/s
= 23 + 47 = 60m/s
At (250.0+B) seconds
250.0+347 =
547 seconds
Distance d,
= 70 x 597
= 41790
It also moves at
(45.0 + c)
= 45 + 19
= 64m/s
Time = 800 + B
= 800 + 347
= 1147
Distance,
= 64 x 1147
= 73408m
Total distance,
= 73408 + 41790
= 115,198
Total time,
= 597 + 1147
= 1744
Average speed,
= Total distance / total time
= 115198/1174
= 66.053m/s
Final answer:
The average speed over the entire time can be calculated by first finding the distances the train travels over both periods, then finding the total distance and the total time, and finally dividing the total distance by the total time. The value must be rounded to three significant figures.
Explanation:
You can find the average speed of the train over the full-time interval by dividing the total distance travelled by the total time. To begin with, you would have to find the distances the train covered during both periods.
- The distance (D1) it travelled during the first period can be found by multiplying the average speed (23.0 + A) by the time (250.0 + B).
- The distance (D2) it travelled during the second period can be calculated by multiplying the average speed (45.0 + C) by the time (800.0 + B).
Then you add D1 and D2 to get the total distance (TD). This will be (D1 + D2). The total time (TT) will be found by adding both time intervals, which means it equals (250.0 + B) + (800.0 + B). You then divide the total distance by the total time to get the average speed, i.e., TD/TT. Lastly, round the average speed to 3 significant figures.
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