Answer: 330.88 J
Explanation:
Given
Linear velocity of the ball, v = 17.1 m/s
Distance from the joint, d = 0.47 m
Moment of inertia, I = 0.5 kgm²
The rotational kinetic energy, KE(rot) of an object is given by
KE(rot) = 1/2Iw²
Also, the angular velocity is given
w = v/r
Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy
w = v/r
w = 17.1 / 0.47
w = 36.38 rad/s
Now, substituting the value of w, with the already given value of I in the equation, we have
KE(rot) = 1/2Iw²
KE(rot) = 1/2 * 0.5 * 36.38²
KE(rot) = 0.25 * 1323.5
KE(rot) = 330.88 J
Answer:
Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.
Explanation:
So, we are given the following set of infomation in the question given above;
=> "spherical Gaussian surface of radius R centered at the origin."
=> " A charge Q is placed inside the sphere."
So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?
The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;
REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.
Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.
Answer:
B) waves speed up
C) waves bend away from the normal
Explanation:
The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:
where
c is the speed of light in a vacuum
v is the speed of light in the medium
We can re-arrange this equation as:
So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is
B) waves speed up
Moreover, when light enters a medium bends according to Snell's Law:
where
are the index of refraction of the 1st and 2nd medium
are the angles made by the incident ray and refracted ray with the normal to the interface
We can rewrite the equation as
So we see that if the index of refraction of the second medium is lower (
), then the ratio 
is larger than 1, so the angle of refraction is larger than the angle of incidence:
This means that the wave will bend away from the normal. So the other correct option is
C) waves bend away from the normal
The free-fall acceleration on the second planet is one-fourth the value of the first planet.
Calculation:
Consider the mass of planet A to be, M
the mass of planet B to be, Mₓ = M
the radius of planet A to be, R₁
the radius of planet B to be, R₂
The acceleration due to gravity on planet A's surface is given as:
g = GM/R₁² - (1)
Similarly, the acceleration due to gravity on planet B's surface is given as:
g' = GM/R₂² [where, R₂ = 2R₁]
= GM/4R₁² -(2)
From equation 1 & 2, we get:
g/g' = GM/R₁² ÷ GM/4R₁²
g/g' = 4/1
Thus we get,
g' = 1/4 g
Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.
Learn more about free-fall here:
<u>brainly.com/question/13299152</u>
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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
