Answer:
A change has occurred in a nucleus
Answer:

Explanation:
Since, as we know, the potential difference 'ΔV' is the difference of between the Potential energy per unit charge U/qo at one point 'B' to Potential energy per unit charge at other point 'A'. It so happens when a test charge 'qo' moves from point A to B, the potential difference becomes the change of potential energy of the system, i.e.
Answer:
c. 70 Ω
Explanation:
The R and R resistors are in parallel. The 2R and 2R resistors are in parallel. The 4R and 4R resistors are in parallel. Each parallel combination is in series with each other. Therefore, the equivalent resistance is:
Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)
Req = R/2 + 2R/2 + 4R/2
Req = 3.5R
Req = 70Ω
Answer:
a) 9.99 s
b) 538 m
c) 20.5 s
d) 1160 m
Explanation:
Given:
x₀ = 0 m
y₀ = 49.0 m
v₀ = 113 m/s
θ = 60.0°
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find t.
vᵧ = aᵧ t + v₀ᵧ
(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)
t ≈ 9.99 s
b) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.
vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)
(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)
y ≈ 538 m
c) When the projectile lands, y = 0 m. Find t.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²
You'll need to solve using quadratic formula:
t ≈ -0.489, 20.5
Since negative time doesn't apply here, t ≈ 20.5 s.
d) When the projectile lands, y = 0 m. Find x. (Use answer from part c).
x = x₀ + v₀ₓ t + ½ aₓ t²
x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²
x ≈ 1160 m
Answer:
10 N.m
Explanation:
Force * Perpendicular distance.
= 20 N * 0.5 m
= 10 N.m