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,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

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,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

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

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

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

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago

The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force

yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given

N = W Cos30

N = (500) Cos30

N = 433 N

3 0

3 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

3 years ago