Answer:
A. 5521.86 secs
B. 7680.65 m/s
Explanation:
Parameters given:
Radius of orbit of the space station, R = 6380000 + 370000 = 6750000 m
Mass of earth = 5.97 * 10^24 kg
Gravitational constant, G = 6.67 * 10^(-11) Nm²/kg²
A. Orbital period, T, can be obtained using the formula:
T²/R³ = (4 * pi²) / (G * M)
T²/(6750000³) = (4 * 3.142²) / (6.67 * 10^(-11) * 5.97 * 10^24)
T² = (4 * 3.142² * 6750000³) / (6.67 * 10^(-11) * 5.97 * 10^24)
T² = 30490944.39
T = 5521.86 secs
B. Orbital velocity, v, can be obtained by using the formula:
v² = (G * M) / R
v² = (6.67 * 10^(-11) * 5.97 * 10^(-24)) / 6750000
v² = 58992384
v = 7680.65 m/s
Answer:
The position for mass is
Explanation:
Let x(t) donate the position of mass at time t,Then x satisfies the differential equation
The general solution is
It follows
thus the position of mass is
My answer -
The condition in the nucleus are likely to result in the nucleus
breaking apart is called the nuclear fission. But if it is the atomic
nucleus that is breaking apart, it should be called as the nuclear
fusion. These things usually can be caused by the decay of the
radioactive and the nuclear reaction.
P.S
Happy to help you have an AWESOME!!! day :)
<span>LAWS OF MOTION:
The motion of an object parallel to the earth's surface is: Horizontal </span>
Answer:
a)
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
Where:
: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ().
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!