Answer:
The length of the object would shrink to zero which is not possible.
Explanation:
A rocket or any body cannot reach the speed of light because according to theory of relativity the and the Lorentz factor the length of the object would shrink to zero and the time dilation for that body would be infinite.
The Lorentz factor is given as:
where:
v = speed of the moving object
c = speed of light
To develop this problem use the concept of the sum given pressure in the tank. At the bottom of the tank the pressure of this will be given by atmospheric pressure, the pressure given by the oil and the pressure by the water, that is to say that mathematically the pressure would be
<em>Note: Here the pressures are expressed in terms of density (), gravity (g) and thickness (t) or height (h). If we rearrange this equation to find the oil thickness we will have to,</em>
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Our values are given as,
Replacing we have that the thickness of the oil is:
Therefore the thick of the oil is 0.5947m
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration.
The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N.
a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r.
The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec.
So a_N = 114 m/sec^2.
g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
Longer I took a little physics but I remember them saying that a small star lasts longer than a larger star
Answer:
D) The capacitor will pull the material into the space between the plates, and the potential energy stored between the plates of the capacitor will decrease by a factor of 3
Explanation:
points to note
A) inserting a material with dielectric constant of 3 between the plates means that the capacitance C will increase by 3.
B) Since the battery is disconnected, the potential difference V between plate will not remain constant.
C) increasing the capacitance reduces the potential difference across the plate, and from C = Q/V it can be seen that the charges on the plate remains constant.
From the proof in the image below, the reduction in the potential energy of the capacitor is due to the energy used by the capacitor to pull the dielectric material into the space.