I thought you were going to ask for the resistance of the unknown
series resistor. Since you only want the equivalent resistance of the
circuit, you don't even need to know the resistance of the lamp.
I = E / R
Current through the circuit = (voltage of the battery) / (circuit resistance).
0.5 = (12) / R
Multiply each side by 'R' : (0.5) R = 12
Multiply each side by 2 : <em>R = 24 ohms</em>
(Since the resistance of the lamp is 10 ohms, the
unknown series resistor is the other 14 ohms.)
True
Explanation:
In order to apply and experience the beauty of mathematics, a good comprehension of the discipline is required.
In fact, this is general to anything in life. To fully maximize any potential, one must be well groomed about what exactly that thing is.
Mathematics is a science that deals with the logic of shapes and numbers. There are different branches of this discipline with real world adoption.
The tenets of mathematics cuts across business, science, arts, e.t.c.
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Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
