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First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
the angular speed is
And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
the angular speed is
And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
Answer:
The earthquake occurred at a distance of 1122 km
Explanation:
Given;
speed of the P wave, v₁ = 8.5 km/s
speed of the S wave, v₂ = 5.5 km/s
The distance traveled by both waves is the same and it is given as;
Δx = v₁t₁ = v₂t₂
let the time taken by the wave with greater speed = t₁
then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.
v₁t₁ = v₂t₂
v₁t₁ = v₂(t₁ + 1.2 min)
v₁t₁ = v₂(t₁ + 72 s)
v₁t₁ = v₂t₁ + 72v₂
v₁t₁ - v₂t₁ = 72v₂
t₁(v₁ - v₂) = 72v₂
The distance traveled is given by;
Δx = v₁t₁
Δx = (8.5)(132)
Δx = 1122 km
Therefore, the earthquake occurred at a distance of 1122 km
Answer:
Jim's kinetic energy is 54.67 J.
Explanation:
Given that,
Mass, m = 15 kg
Velocity, V = 2.7 m/s
We need to find the Jim's kinetic energy. We know that when the object is in motion, it has kinetic energy. This energy is given by :
E = 54.67 J
So, Jim's kinetic energy is 54.67 J. Hence, this is the required solution.