Answer: (a) t1 = omega1/alpha
(b) theta1 = 1/2 * alpha*theta1^2
(c) t2 = omega2/5*alpha
Explanation: see attachment
The people in a location in Florida have reduced their water usage drastically since the last few months. People are taking shor
2 answers:
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Complete Question
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300
. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Answer:
a
The additional resistance is
b
The rate at which internal energy increase at the supply is
c
The rate at which internal energy increase in the battery is
d
The rate at which internal energy increase in the added series resistance is
e
the increase rate of the chemically energy in the battery is
Explanation:
From the question we are told that
The open circuit voltage is
The internal resistance is
The emf of each battery is
The internal resistance of the battery is
The charging current is
Let assume the the additional resistance to to added to the circuit is
So this implies that
The total resistance in the circuit is
Substituting values
And the difference in potential in the circuit is
=>
Now according to ohm's law
Substituting values
Making the subject of the formula
So
The increase rate of internal energy at the supply is mathematically represented as
Substituting values
The increase rate of internal energy at the batteries is mathematically represented as
Substituting values
The increase rate of internal energy at the added series resistance is mathematically represented as
Substituting values
Generally the increase rate of the chemically energy in the battery is mathematically represented as
Substituting values
Answer:
2.03 x 10²⁴N
Explanation:
Given parameters:
Mass of moon = 7.34 x 10²²kg
Mass of the earth = 5.97 x 10²⁴kg
Distance = 3.8 x 10⁵km
Unknown:
Gravitational force of attraction = ?
Solution:
To find the gravitational force of attraction between the masses, we use the expression below;
F =
G is the universal gravitation constant
m is the mass
1 and 2 represents moon and earth
r is the distance
F =
F = = 2.03 x 10²⁴N