B, larceny because that's theft of personal property.
Answer:
Answer B
Explanation:
An increase in resistance makes it harder for the electric current to pass through
Answer:
Explanation:
Given
Resistor A has length 
and Resistor B has Length 
and Resistance is given by

Considering
and A to be constant thus
because 
(a)When they are connected in series
As the current in series is same and power is 
therefore
as R is greater for second resistor
(b)if they are connected in Parallel
In Parallel connection Voltage is same

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2
Answer:

Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
