A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
1 answer:
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Answer:
V = 10.88 m/s
Explanation:
V_i =initial velocity = 0m/s
a= acceleration= gsinθ-cosθ
putting values we get
a= 9.8sin25-0.2cos25= 2.4 m/s^2
v_f= final velocity and d= displacement along the inclined plane = 10.4 m
using the equation
v_f= 7.04 m/s
let the speed just before she lands be "V"
using conservation of energy
KE + PE at the edge of cliff = KE at bottom of cliff
(0.5) m V_f^2 + mgh = (0.5) m V^2
V^2 = V_f^2 + 2gh
V^2 = 7.04^2 + 2 x 9.8 x 3.5
V = 10.88 m/s
Answer:
The volume of water displaced by the raft is 0.233 m³
Explanation:
The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid
The given parameters are;
The combined mass of the person and the raft, m = 233 kg
The liquid on which the raft is located = Water
The density of water, = 1000 kg/m³
Weight = Mass, m × g
Where;
m = The mass of the object
g = The acceleration due to gravity = 9.8 m/s²
Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg
The combined weight of the person and the raft, = 233 kg × 9.8 m/s² = 2,283.4 N
Therefore;
The buoyant force = 2,283.4 N = The weight of the water displaced
The mass of the water displaced, , = 2,283.4 N/(9.8 m/s²) = 233 kg
Density = Mass/Volume
The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.