All categories

3 years ago

A walkman uses four standard 1.5 V batteries. How much resistance is in the circuit if it uses a current of 0.02A? *

Physics

1 answer:

hodyreva [135]3 years ago

7 0

Answer:

75ohms

Explanation:

V= IR

V = 1.5volts

I = 0.02A

1.5 = 0.02×R

Making R the subject

R = 1.5/0.02

R = 75ohms

The resistance in the circuit will be 75ohms

You might be interested in

A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction

lana66690 [7]

When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
$\theta_c = \arcsin ( \frac{n_r}{n_i} )$ (2)
where $n_r$ is the refractive index of the second medium and $n_i$ is the refractive index of the first medium.

We can find the ratio $n_r / n_i$ by using Snell's law:
$n_i \sin \theta_i = n_r \sin \theta_r$ (1)
where
$\theta_i$ is the angle of incidence
$\theta_r$ is the angle of refraction

By using the data of the problem and re-arranging (1), we find
$\frac{n_r}{n_i} = \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566$

and if we use eq.(2) we can now find the value of the critical angle:
$\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}$

3 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

lianna [129]

<h2>Answer:</h2>

<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>

<h2>Explanation:</h2>

In the question,

Let us say the height from which the arrow was shot = h

Distance traveled by the arrow in horizontal = 61 m

Angle made by the arrow with the ground = 2°

So,

From the <u>equations of the motion</u>,

$61 =u.t\\t=\frac{61}{u}$

Now,

Also,

Finally, the angle made is 2 degrees with the horizontal.

So,

Final horizontal velocity = v.cos20°

Final vertical velocity = v.sin20°

Now,

u = v.cos20° (No acceleration in horizontal)

Also,

$v=u+at\\vsin20=0+9.8(t)\\t=\frac{v.sin20}{9.8}$

So,

We can say that,

$\frac{v.sin20}{9.8}=\frac{61}{v.cos20}\\v^{2}.sin20.cos20=597.8\\v^{2}=1860.56\\v=43.13\,m/s$

<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>

5 0

3 years ago

Nuclear Energy definition

Svetlanka [38]

Answer:

The energy released during nucleur fissionor fusion , espicially when used to generate

Explanation:

Distionary.

8 0

2 years ago

When the voltage across a steady resistance is doubled, the current?

natima [27]

I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.</span>

The next question is what I'm not sure about:

Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).

What I did initially was: P = IV = (100mA)(2V) = 2 W

But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

8 0

3 years ago

A transformer has input voltage and current of 12 V and 4 A

Advocard [28]

Explanation:

It is given that,

Input voltage, $V_i=12\ V$