Answer:
14.7 m/s.
Explanation:
From the question given above, the following data were obtained:
Time (t) = 1.5 s
Acceleration due to gravity (g) = 9.8 m/s².
Height = 11.025 m
Final velocity (v) = 0 m/s
Initial velocity (u) =?
We, can obtain the initial velocity of the penny as follow:
H = ½(v + u) t
11.025 = ½ (0 + u) × 1.5
11.025 = ½ × u × 1.5
11.025 = u × 0.75
Divide both side by 0.75
u = 11.025/0.75
u = 14.7 m/s
Therefore, the penny was travelling at 14.7 m/s before hitting the ground.
Answer:
T=26.03 N
Explanation:
Given that
Distance between poles = 12 m
Mass of block m= 4 kg
Sag distance = 5 m
Lets take tension in the clothesline is T.
The component of tension in vertical direction will be T cosθ.
By force balancing
2 T cosθ = 40
here 
θ=39.80°
2 T cos39.8 = 40
T=26.03 N
Answer:
N = 337.96 N
Explanation:
∅ = 32º
F = 249 N
m = 21 Kg
N = ?
We can apply:
∑ F = 0 (↑)
- Fy - W + N = 0 ⇒ N = Fy + W
⇒ F*Sin ∅ + m*g = N
⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)
⇒ N = 337.96 N (↑)
Answer:
The zero field location has to be on the line running between the two point charges because that's the only place where the field vectors could point in exactly opposite directions. It can't be between the two opposite charges because there the field vectors from both charges point toward the negative charge.
