This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute
Answer:
Potential energy of book = 7.5 J
Explanation:
Given:
Weight of book = 5 N
Height of shelf = 1.5 meter
Find:
Potential energy of book
Computation:
Weight = Mass x Acceleration of gravity
Mass x Acceleration of gravity = 5 N
Potential energy = Mass x Acceleration of gravity x Height
Potential energy of book = Mass x Acceleration of gravity x Height
We know that;
Mass x Acceleration of gravity = 5 N
So,
Potential energy of book = 5 x 1.5
Potential energy of book = 7.5 J
Answer:
false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy
Explanation:
mechanical energy = potential energy + kinetic energy = constant
differentiating both side
Δ potential energy + Δ kinetic energy = 0
Δ potential energy = - Δ kinetic energy
first statement is true.
Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows
change in gravitational potential energy = change in kinetic energy + work done against friction .
So given 2 nd option is incorrect.
In case of no change in gravitational energy , work done is equal to
change in kinetic energy.
Answer:
t = 1.05 s
Explanation:
Given,
The distance between your vehicle and car, 100 ft
The constant speed of your vehicle, u = 95 ft/s
Since, the velocity is constant, a =0
If the car stopped suddenly, time left for you to hit the brake, t = ?
Using the second equation of motion,
S = ut + ½ at²
Substituting the given values in the equation
100 = 95 x t
t = 100/95
= 1.05 s
Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s
