Ask question

Ask question

All categories

3 years ago

15

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc

reasing at a rate of 3.0 J/s, how fast is the current changing

1 answer:

lys-0071 [83]3 years ago

5 0

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

You might be interested in

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?

jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina ( $Q$ ), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

$Q = \frac{\pi}{4}\cdot D^{2}\cdot v$ (1)

Donde:

$D$ - Diámetro de la manguera, medido en pies.

$v$ - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que $D = \frac{1}{12}\,ft$ y $v = 5\,\frac{ft}{s }$ , entonces el gasto de gasolina es:

$Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)$

$Q \approx 0.0273\,\frac{ft^{3}}{s}$

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0

3 years ago

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

now, calculating at the angular velocity

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

$\omega =\dfrac{2\pi}{1.25}$

ω = 5.026 rad/s

now, radial acceleration

a = r ω²

a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0

3 years ago

How to calculate displacement?

Blizzard [7]

-- Take a straight ruler.

-- Lay it down with the 'zero' mark at the start point.

-- Rotate it around the start point until the end point is also touching the edge of the ruler.

-- From the marks on the ruler, read the straight-line distance from the start point to the end point.

-- Without moving the ruler, observe and write down the DIRECTION from the start point to the end point.

-- The Displacement is the straight-line distance and direction from the start point to the end point.

4 0

3 years ago

A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance. A man pushes a

skelet666 [1.2K]

Answer:

B.) a wheel and axle and a lever

Explanation:

P.S - The exact question is -

Given - A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance. A man pushes a wheelbarrow.

To find - Which simple machines make up a wheelbarrow?

A.) a pulley and an inclined plane

B.) a wheel and axle and a lever

C.) a pulley and a wheel and axle

D.) a lever and a wedge

Proof -

The correct option is - B.) a wheel and axle and a lever

Wheelbarrows are used to carry more goods from place to place by using minimal force as compared when goods are carried by hand.

With this machine, During hauling people can save time.

4 0

2 years ago

Read 2 more answers