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3 years ago

A 4-A current is maintained in a simple circuit with a total resistance of 2 Ω. How much energy is dissipated in 3 seconds?

Physics

1 answer:

zvonat [6]3 years ago

8 0

Answer: Energy dissipated E = 96J

Explanation:

Given:

Current I = 4A

Resistance R = 2 Ohms

Time t = 3 seconds

The energy dissipation in an electric circuit can be derived from the equation below:

E = IVt ....1

Where;

I = current, V = Voltage (potential difference), t= time and E = energy dissipated

But we know that;

V = I×R .....2

Substituting equation 2 to 1, we have

E = IVt = I(I×R)t = I^2(Rt)

Substituting the values of I,R and t

E = 4^2 × 2 ×3 = 96J

Energy dissipated E = 96J

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What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.0

SOVA2 [1]

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

$Energy=h\times frequency$

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So, Formula for energy:

$Energy=h\times \frac {c}{\lambda}$

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

$2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}$

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

<u>Wavelength = 736.67 nm</u>

8 0

3 years ago

7. How is the current found in a parallel circuit?

Naya [18.7K]

Answer:

the current in a parallel circuit is found by idk

Explanation:

4 0

3 years ago

A taxi is travelling at 15m/s. Its driver accelerates with acceleration 3m/s^2 for 4 s. What would be the new velocity?..... pls

Aleksandr-060686 [28]

Answer:

27 m/s

Explanation:

Given:

v₀ = 15 m/s

a = 3 m/s²

t = 4 s

Find: v

v = at + v₀

v = (3 m/s²) (4 s) + (15 m/s)

v = 27 m/s

5 0

3 years ago

Read 2 more answers

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

4 years ago

A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re

White raven [17]

A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law.

$I= \frac{V}{R}$

Where:
I = current (A)
V = Voltage (V) (emf)
R = Resitance (Ω)

Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
$V = IR$

However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

$\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}$

$\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059$
$\frac{1}{R_{T}} =0.0613$

Get the reciprocal of both sides and divide:

$R_{T} = \frac{1}{0.0613} =16.32$

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
$V = IR$
$V = (1.8)(16.32)$
$V = 29.376V$

The emf of the battery is 29.376V

B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{65}$
$I= 0.45A$

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{25}$
$I= 1.18A$
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{170}$
$I= 0.17A$

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.

4 0

3 years ago