Answer:
30.96 m
Explanation:
If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:
d = v * Δt
v = 0.8*c = 0.8 * 3e8 = 2.4e8
Δt = ζ = 129 ns = 1.29e-7 s
d = 2.4e8 * 1.29e-7 = 30.96 m
This is the distance as measured by observer A.
Answer:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
1.6 m/s west is the answer
Answer:
vₓ = xg/2y
Explanation:
In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.
By newton equation of motion we know that
y = v₀ t - ½ g t²
t = 2y / g
This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance
vₓ = x/t
vₓ = xg/2y
vₓ = xg/2y
Where we assume that x and y are known.
Answer: 0.62
Explanation:
Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).
Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R
Since W = mg, W = 38.4 ×10
W= 384N =R
Normal reaction = 384N
The horizontal force acting on the body will be the moving force which is 238N
Coefficient of friction = Fm/R
Coefficient of friction = 238/384
Coefficient of friction = 0.62
Therefore, coefficient of kinetic friction between the box and the floor is 0.62
