Answers
Answer:
Explanation:
given,
time taken to complete the each orbit = 144 minutes
t = 144 x 60 = 8640 s
mass of the earth = 5.98 x 10²⁴ Kg
radius of earth,R = 6.38 x 10⁶ m
Using Kepler's 3rd law
r = 9.1 x 10⁶ m
the altitude of the satellite
H = r - R
H = 9.1 x 10⁶ - 6.38 x 10⁶
H = 2.72 x 10⁶ m
Related Questions
A satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation? A. The resultant force is the same direction as the satellite’s acceleration. B. The gravitational force acting on the satellite is negligible. C. There is no resultant force on the satellite relative to the Earth. D. The satellite does not exert any force on the Earth.
Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).
You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling fan with light kits be installed in five different rooms. Each fan contains a light kit that can accommodate four 60-watt lamps. Each fan motor draws a current of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than three hours at a time and therefore must be considered a continuous-duty device. The fans are to be connected to a 15-ampere circuit. Because the devices are continuous-duty, the circuit current must be limited to 80% of the continuous connected load. How many fans can be connected to a single 15-ampere circuit
In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.
Answers
Answer:
Explanation:
Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.
This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.
So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.
Answers
Answer:
B
Explanation:
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?
Answers
Answer:
4500 J and 3000 J
Explanation:
According to conservation of momentum
Given that m_2 = 1.5 m_1 , so
the kinetic energy of each piece is
substituting the value of V1 in the above equation
Given that
K_1 + k_2 = 7500 J
1.5 K_2 + K_2 = 7500
K_2 = 7500 / 2.5
= 3000 J
this is the KE of heavier mass
K_1 = 7500 - 3000 = 4500 J
this is the KE of lighter mass
Final answer:
The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.
Explanation:
The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.
Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.
Learn more about Kinetic Energy Acquisition here:
#SPJ3
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
Answers
Incomplete question.The complete question is here
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.
Answer:
F= 0.034 N
Explanation:
Given Data
Outer=2 cm
Edge of blade=21 cm
Mass=7.7 g
Length of blade=21 cm
The last 2cm is treated as if they were at a point 20cm from the center of rotation
To Find
Force=?
Solution
Convert the given frequency to angular frequency
ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)
ω= 3/2*π rad/sec
Now to find centripetal force.
F = m×v²/r
F= m×ω²×r
Put the data
F = 0.0077 kg × (3/2×π rad/sec)²× 0.20 m
F= 0.034 N
Answers
The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.
What is angular velocity?
The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.
ω = θ t
Where,
θ is the angle of rotation,
tis the time
ω is the angular velocity
The given data in the problem is;
u is the initialvelocity=0
α is the angularacceleration = 4.0 rad/s²
t is the time period=
n is the number of revolution = 10 rev
From Newton's second equation of motion in terms of angular velocity;
Hence the angular velocity will be 22.41 rad/s.
To learn more about angularvelocity refer to the link
Answer:
= 22.41rad/s
Explanation:
First, we know that:
a = 4 rad/s^2
S = 10 rev = 62.83 rad
Now we know that:
where is the final angular velocity,
the initial angular velocity, a is the angular aceleration and S the radians.
Replacing, we get:
Finally, solving for :
= 22.41rad/s