Ask question

Ask question

All categories

3 years ago

13

What voltage indicates the voltmeter connected to the ends of a conductor, if the ammeter connected in series with this conducto

r indicates a current of 400 mA. The resistance of the conductor is equal to 5kΩ.

1 answer:

ycow [4]3 years ago

3 0

Explanation:

V= Current × Resistance

V= 400mA × 5kohms

V= 2000V

You might be interested in

i know this is a lot to do but i will give you the highest amount of points (100, but you get 50 from it.) and i will cash app y

victus00 [196]

Answer:

mel mel 4 ever

Explanation:

5 0

4 years ago

Find the velocity v=ds/dt at t=3

IRINA_888 [86]

Answer:

you need at least two out of the three to get any aenser

7 0

3 years ago

A vector quantity must include both magnitude and direction. Which measurement is a vector quantity? A) the rain accumulation at

bogdanovich [222]

Answer:

B)The motion of water in an ocean current

Explanation:

With respect to measurements, a vector has both a magnitude and a direction. The first three examples (maximum height of a hill, air temperature, and rain accumulation) are magnitudes only. The fourth example (motion of water in an ocean current) is a vector, because it has a magnitude (speed) and a direction (with the current).

4 0

3 years ago

The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale

Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

6 0

2 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago