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2 years ago

. A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is

Physics

2 answers:

Kobotan [32]2 years ago

8 0

Answer:

Explanation:

Given that,

A light bulb has a resistance of 2.9ohms

R = 2.9 ohms

And a battery of 1.5V is applied

V = 1.5 V

We want to find the rate of energy transformed

First we need to know what rate of energy is

Rate of energy implies that we want to find power. Power is the rate at which work is done

P = Workdone / time

Then,

In electronic, the power dissipated by a resistor is given as

P = V² / R

P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

So, the rate at which electrical energy transformed in the lightbulb is 0.776 Watts

e-lub [12.9K]2 years ago

7 0

Answer:

The rate of electrical energy transformed in the light bulb is 0.78 Joules per seconds.

Explanation:

Electrical energy is given by,

E = IVt

The rate of electrical energy transformed in the light bulb can be determined by;

$\frac{E}{t}$ = IV

From Ohms law which states that the amount of current passing through a metallic conductor, e.g wire, is directly proportional to the potential difference across its ends, provided that the temperature is constant.

i.e V = IR

where: V is the potential difference, I is the current and R is the resistance of the conductor

So that: R = 2.9 Ohms and V = 1.5 Volt, find I.

⇒ I = $\frac{V}{R}$

= $\frac{1.5}{2.9}$

= 0.5172

I = 0.52 A

Then;

$\frac{E}{t}$ = IV

= 0.52 × 1.5

= 0.78 J/s

The rate of electrical energy transformed in the light bulb is 0.78 Joules per seconds.

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The trough of the sine curve used to represent a sound wave corresponds to

iren [92.7K]

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

4 0

2 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

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

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

2 years ago

Compare the benefits of wildfires to grasslands, northern forests, and deciduous forests.

Dimas [21]

Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.

6 0

2 years ago

Read 2 more answers

What is the electric potential vtot at the center of the square? make the usual assumption that the potential tends to zero far

Romashka-Z-Leto [24]

Missing figure of the problem: http://tsephysics.weebly.com/uploads/5/1/9/3/51934203/477140_orig.jpg

Solution:
Assuming the potential is zero at infinite distance from the charge, then the potential at a certain distance r from a single point charge is
$V(r)=k_e \frac{q}{r}$
where $k_e=8.99\cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant.

In our problem, we just have to superimpose the potential generated by every charge. The diagonal of the square is $\sqrt{2} d$ , therefore the distance between each charge and the center of the square is $\frac{ \sqrt{2} }{2} d$ .
So, the total potential is:
$V=V_1+V_2+V_3+V_4=$
$=k_e \frac{q}{ \frac{ \sqrt{2}d }{2} }+ k_e \frac{2q}{ \frac{ \sqrt{2}d }{2} }+k_e \frac{5q}{ \frac{ \sqrt{2}d }{2} }-k_e \frac{3q}{ \frac{ \sqrt{2}d }{2} }=$
$=5 \sqrt{2} k_e \frac{q}{d}$

7 0

2 years ago

In the Bohr model of the hydrogen atom, an electron (massm=9.1×10−31kg) orbits a proton at a distance of 5.3×10−11m. The proton

Marysya12 [62]

Answer:

$\omega = 6.557 \times 10^{16}\ rev/s$