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Tanzania [10]
3 years ago
7

Two identical resistors are connected in parallel across a 26-V battery, which supplies them with a total power of 7.1 W. While

the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged. Find (a) the initial resistance of each resistor, and (b) the total power delivered to the resistors after one resistor has been heated.
Physics
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

A) R = 190.42 Ω

B) P = 5.325 W

Explanation:

We are given;

Total power;P_tot = 7.1 W

Voltage;V = 26 V

A)We are told that while the battery is still connected, one of the resistors is heated, so that its resistance doubles.

Thus, the power is doubled.

Now, formula for power is;

P = IV

Thus, since power is doubled, we have;

P = 2(IV)

Now, formula for current is; I = V/R

So, P = 2V²/R

Making R the subject, we have;

R = 2V²/P

In this question, P is p_total = 7.1 W

Thus;

R = (2 × 26²)/7.1

R = 190.42 Ω

B) Now, the resistance of the resistors are R and 2R.

Formula for power in this context is;

P = V²/R

Thus,

Total power delivered to the resistors is;

P = V²/R + V²/2R

P = 3V²/2R

P = (3 × 26²)/(2 × 190.42)

P = 5.325 W

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tensa zangetsu [6.8K]

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

7 0
3 years ago
How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0
2 years ago
If V is the original volume, V' is the new volume, T is the original Kelvin temperature, and T' is the new Kelvin temperature, h
madam [21]

Answer:

\frac{V}{T} = \frac{V'}{T'}

Explanation:

Given the following data;

Original volume = V

New volume = V'

Original temperature = T

New temperature = T'

To write an expression for Charles's law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles law is given by the formula;

\frac {V}{T} = K

\frac{V}{T} = \frac{V'}{T'}

4 0
3 years ago
Which law of physics relates electric fields and current
riadik2000 [5.3K]

Answer:

Ohms law

Explanation:

Which states that the current flowing through any cross-section of the conductor is directly proportional to the potential differenceapplied across its end, provided physical conditions like temperature and pressure remain constant.

7 0
3 years ago
Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ?0, b
cestrela7 [59]

Answer:

Explanation:

Moment of inertia of a disc = 1/2 M R²

Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and   - ω₀ .

We shall apply law of conservation of angular momentum .

initial total angular momentum

I x ω₀ - 4I x ω₀ = - 3Iω₀

Let final common angular momentum be ω

total final angular momentum = ( I + 4I ) ω

Applying law of conservation of angular momentum

( I + 4I ) ω =  - 3Iω₀

ω = - 3 / 5 ω₀ .

b )

Initial total rotational K E

= 1/2 I ω₀² + 1/2 4I ω₀²

= 1/2 x5I ω₀²

Final total rotational K E

= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²

= 1/2 x 9 / 5 I ω₀²

= 9 / 10I ω₀²

change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²

(9/10 - 5/2) xI ω₀²

=( .9 - 2.5 )I ω₀²

= - 1.6 I ω₀² Ans

8 0
3 years ago
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