Ask question

Ask question

All categories

3 years ago

10

A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 200 mA. What is the

current input?

1 answer:

goldfiish [28.3K]3 years ago

4 0

Answer:

20 mA

Explanation:

We have given that the transformer convert 120 V to 12 V means it is a step down transformer

The turn ratio of the transformer is given as $a=\frac{N_1}{N_2}=\frac{V_1}{V_2}=\frac{120}{12}=10$ here $V_!$ is primary voltage and $V_2$ is secondary voltage

As the voltage decreases from primary to secondary so current will increase from primary to secondary

We have given output current that is secondary current so primary that is input current will be less

So input current $=\frac{200}{10}=20mA$

You might be interested in

Heat is transfered from the heating elements to the pot

Varvara68 [4.7K]

Hear is transferred from the heating elements to the Pot by Conductivity

8 0

3 years ago

What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass

Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

$F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz$

The frequency of the simple pendulum is given as;

$F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m$

Thus, the length of the simple pendulum is 0.25 m.

8 0

2 years ago

Podrian ayudarme? Llevo dias intentando pero no se como 1. el profesor no explico

Paraphin [41]

You have to upload this in the area of mathematicians..!

8 0

3 years ago

A node is a point on a standing wave that has no displacement from the rest position. At the nodes, _____.

Kitty [74]

Answer: (B) There is complete destructive interference between the incoming and reflected waves

Explanation:

For example, if you pluck a guitar the waves will travel back and forth. They consist of nodes and anti-nodes. It is created, when the wave traveling to one side and bounces of the other end and comes back. As it travels to the other side, it is reflected thus, comes back. So standing waves occurs when there is interference.

When the wave is produced, the points where the string is not moving are called nodes and where they are moving are called anti-nodes. The positions where nodes are produced, destructive interference occurs and where anti-nodes are produced, constructive interference occurs

8 0

2 years ago

Read 2 more answers

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago