Answer:
354.72 m/s
Explanation:
= mass of lead bullet
= specific heat of lead = 128 J/(kg °C)
= Latent heat of fusion of lead = 24500 J/kg
= initial temperature = 27.4 °C
= final temperature = melting point of lead = 327.5 °C
= Speed of lead bullet
Using conservation of energy
Kinetic energy of bullet = Heat required for change of temperature + Heat of melting
Answer:
0.0006091222 m
Explanation:
q = Charge = 42 pC
V = Voltage = 620 V
= Permittivity of free space =
Electric potential is given by (at r = R)
The radius of the drop is 0.0006091222 m
Answer:
The length of the flagpole is approximately 87.43 m
Explanation:
The given parameters of the cable attached to the flagpole are;
The point along the flagpole at which the cable is attached = 32.0 m
The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°
The downward force exerted by the cable, = 1.233 × 10⁴ N
The force exerted by the cable to the left = 1.233 × 10⁴ N
Let 'W' represent the weight of the flagpole, at equilibrium, we have;
The sum of vertical forces = 0
Therefore;
+ W - R = 0
W - R = -1.233 × 10⁴ N
Taking moment about the support at the base of the pole, we get;
1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0
∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d
W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N
R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373
R ≈ 30,382.373 N
Taking moment about the point of attachment of the cable to the ground, we have;
W × ((d/2) × cos(60.0°) + 32) = R × 32
∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281
d = 2 × 43.71281 ≈ 87.43
The length of the flagpole, d ≈ 87.43 m