Answer:
7.41 x 10^-10 J
Explanation:
m = 4.5 kg
M = 6.3 kg
d = 1.7 m
The formula for the gravitational potential energy is given by
Where, G is the Universal gravitational constant
G = 6.67 x 10^-11 Nm^2/kg^2
U = - 1.112 x 10^-9 J
Now the separation is tripled, d' = 3 x d = 3 x 1.7 m = 5.1 m
Let the potential energy is U'
The formula for the gravitational potential energy is given by
Where, G is the Universal gravitational constant
G = 6.67 x 10^-11 Nm^2/kg^2
U' = - 3.71 x 10^-10 J
the work done is equal to the change in potential energy
W = U' - U
W = - 3.71 x 10^-10 + 1.112 x 10^-9
W = 7.41 x 10^-10 J
Density is mass per unit volume,
so to get the volume, divide the mass by the density,
So the volume is
Answer:
1: toward the normal
2:away from normal
3: red because the index of refraction ( the ratio of the speed of light in a vacuum to the speed of light in a material) is increased for the slower moving waves
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
