Answer:
a) I = 13.04 A
b) R = 8.82 ohms
c) 1291.87 kilocalories are generated an hour.
Explanation:
let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.
a) we know that:
P = I×V
I = P/V
= (1500)/(115)
= 13.04 A
Therefore, the current of the heater is 13.04 A
b) we now have voltage and current, according to Ohm's law:
R = V/I
= (115)/(13.04)
= 8.82 ohms
Therefore, the resistance of the heating coil is 8.82 ohms.
c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:
E = P×t
= (1500)(1×60×60)
= 5400000 J
since 1 calorie = 4.81 J
1 kilocalorie = 0.001 calories
E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories
Therefore, 1291.87 kilocalories are produced/generated in one hour.
Answer: B
Explanation:
Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.
Answer:
Explanation:
Given
Bianca is at 
i.e. distance between origin and Bianca is 
time taken to reach Bianca eyes is
i.e. Cracker exploded at 
because it is observed at 
Time taken by second cracker flash to reach Bianca eyes
Therefore it will be observed at 
Answer:
1) in metal object heat transfer through the conduction .In vacuum heat transfer through only radiation .
In only gaseous state or in liquid state the heat transfer through convection hence option D is correct
Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
