Answer:
274N 0.41
Explanation:
As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.
then
<em>friction=mass x gravity x sin(21)</em>
Fr=78kg x 9.8m/s2 x sin(21)=274N
<em>friction= coefficient of kinetic friction x normal force of from the slope</em>
Fr= u x 78kg x 9.8m/s2 x cos(21)=274N
Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
The center of the ghole is the singularity. or a point where extremely large amounts of matter are crushed into an infinitely small amount of space. hope it helps :)
Answer:
K_{e} = 2.0 J
Explanation:
In this exercise you are asked to calculate the elastic potential energy of a spring
= ½ k x²
where k is the spring constant and x is the displacement from equilibrium position
In this exercise, indicate that the spring constant is k = 100 N/m, the length at rest is x₀ = 20 cm = 0.20 m, up to the position x₁ = 40 cm = 0.40 m, therefore the elongation
Δx = x₁ - x₀
Δx = 0.40 - 0.20
Δx = 0.20 m
let's calculate the elastic potential energy
K_{e} = ½ 100 0.20²
K_{e} = 2.0 J
Answer:
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