Answer:
Velocity, quantity that designates how fast and in what direction a point is moving.
Explanation:
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,


Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity



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,


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



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



Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
Answer:
v=12.5 i + 12.5 j m/s
Explanation:
Given that
m₁=m₂ = m
m₃ = 2 m
Given that speed of the two pieces
u₁=- 25 j m/s
u₂ =- 25 i m/s
Lets take the speed of the third mass = v m/s
From linear momentum conservation
Pi= Pf
0 = m₁u₁+m₂u₂ + m₃ v
0 = -25 j m - 25 i m + 2 m v
2 v=25 j + 25 i m/s
v=12.5 i + 12.5 j m/s
Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s
Answer:
0.021 V
Explanation:
The average induced emf (E) can be calculated usgin the Faraday's Law:
<u>Where:</u>
<em>N = is the number of turns = 1 </em>
<em>ΔΦ = ΔB*A </em>
<em>Δt = is the time = 0.3 s </em>
<em>A = is the loop of wire area = πr² = πd²/4 </em>
<em>ΔB: is the magnetic field = (0 - 1.04) T </em>
Hence the average induced emf is:
Therefore, the average induced emf is 0.021 V.
I hope it helps you!
Answer: D) It should be handled in a fume hood, away from open flames.