According to the law of conservation of mass, which of these is true.

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . 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 $\Omega$ . 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 $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

6 0

3 years ago

A person's prescription for bifocals is -0.25 diopter for distant vision and +2.50 diopters for near vision, the near vision bei

Darina [25.2K]

Answer:

net power is + 2.25 D

Explanation:

Given data

distance vision = -0.25 D

near vision = + 2.50 D

to find out

net power

solution

we have given a person lens power for near is - 0.25 diopter and lens power for near power is +2.50 diopter so

net power is sum of both the power vision

so

net power = distance + near power

put both value we get net power

net power = ( -0.25 D) + ( + 2.50 D)

net power = + 2.25 D

so net power is + 2.25 D

6 0

3 years ago

The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t

Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

$B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}$

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

$r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\$

8 0

2 years ago

Why concave lens is called diverging lens

dangina [55]

When a parallel beam of light passes through a convex lens, the rays become farther from one another when the come out. This process of rays is called ''to diverge''. The concave lens makes rays of light diverge, so it is called diverging lens.

6 0

2 years ago

Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to

tamaranim1 [39]

Answer:

Explanation:

A

35 N Small Dog <=======BONE=========> Bigger Dog 42 N

B

Fnet = Large Dog - small dog The forces are subtracted because they are acting in opposite Directions.

Fnet = 42 - 35

Fnet = 7 N

C

m = 2.5 kg

F = 7 N

a = ?

F = m * a

7 = 2.5 a

a = 7 / 2.5

a = 2.8 m/s^2

4 0

1 year ago