Ask question

Ask question

All categories

4 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.

1 answer:

Gnom [1K]4 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

You might be interested in

What other experiments can you do with bubble gum?

Elanso [62]

What happens when you leave it in differnt soda pops.

7 0

3 years ago

PSYCHOLOGY <br> Which of these statements are false ?

nataly862011 [7]

Answer:

The false statement has to be that neruotransmitters are in the spinal cord.

Explanation:

Neruotransmitters are in no was associated with the spinal cord. That is more related to nerves and muscles. Neurotransmitters are in the Phasma Membrane acording to sciencedirect.com.

4 0

2 years ago

While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r

sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m$

If f = 78 kHz,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m$

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

6 0

3 years ago

Describe three physical changes that occur in nature?

Alex Ar [27]

Answer: Physical changes in nature could then be erosion in a mountain, the melting of snow, and a river freezing over from the cold. Since none of these changes affect the chemical composition of the mountain, the snow, or the river, they are physical changes.

Explanation:

8 0

3 years ago

Read 2 more answers

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago