Gravitational potential energy can be calculated using the formula:
Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height
On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.
So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m
So we put that in and solve it.
Explanation:
Formula for angle subtended at the center of the circular arc is as follows.
where, S = length of the rod
r = radius
Putting the given values into the above formula as follows.
= 
= 
= 
Now, we will calculate the charge density as follows.
= 
= 
Now, at the center of arc we will calculate the electric field as follows.
E = 
= 
= 34.08 N/C
Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.
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:
h = 16.67m
Explanation:
If the kinetic energy of the cylinder is 510J:
Where the inertia is given by:
Replacing this value:
Speed of the block will therefore be:
By conservation of energy:
Eo = Ef
Eo = 0
So,
Solving for h we get:
h=16.67m
Answer:
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Explanation:
