Explanation:
Resistors connected in series obey the following equation:
Resistors connected in parallel obey the following equation:
The total current of the circuit will obey the Ohm's Law: V = IR. And the current will be divided across the resistors (bulbs) depending on their resistances. So, if a bulb has a higher resistance, then its current will be lesser, and it will be less bright. If a bulb has a lower resistance, then its current will be higher, and it will be brighter.
According to the above resistances connected in series and parallel, clearly, the resistances (bulbs) connected in series will have more resistance in total, and therefore less current will flow across them, and they will be less bright.
Answer:
a) Block 1 = 72.9kgm/s
Block 2 = 0kgm/s
b) vf = 1.31m/s
c) ∆KE = 936.36Joules
Explanation:
a) Momentum = mass× velocity
For block 1:
Momentum = 2.7×27
= 72.9kgm/s
For block 2:
Momentum = 53(0) (body is initially at rest)
= 0kgm/s
b) Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the block
u1 and u2 are their initial velocity
v is the common velocity
Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)
2.7(27)+53(0) = (2.7+53)v
72.9 = 55.7v
V = 72.9/55.7
Vf = 1.31m/s
c) kinetic energy = 1/2mv²
Kinetic energy of block 1 = 1/2×2.7(27)²
= 984.15Joules
Kinetic energy of block 2 before collision = 0kgm/s
Total KE before collision = 984.15Joules
Kinetic energy after collision = 1/2(2.7+53)1.31²
= 1/2×55.7×1.31²
= 47.79Joules
∆KE = 984.15-47.79
∆KE = 936.36Joules
Answer:
A. Indians who were part of the East India Company’s private army.
Explanation:
The sepoys were Indian soldiers who were recruited into the Company's army. Just before the rebellion, there were over 300,000 sepoys in the army, compared to about 50,000 British. The forces were divided into three presidency armies: Bombay, Madras, and Bengal.
