A man walks 23 km east and then

An iv curve is generated for two ohmic devices, one is a good conductor and the other is a good insulator. which one will have a

allochka39001 [22]

As per the question there are two ohmic devices.

The first ohmic device is a good conductor and the second one is a good insulator.

As per Ohm's law the current flowing through a conductor is directly proportional to the potential difference maintained across the two ends of a conductor at constant temperature and pressure. If V is the potential and I is the current flow through these Ohmic devices,then mathematically it can be written as-

V∝ I [ at constant temperature and pressure]

⇒V= IR

Here R is the proportionality constant called resistance of the material.

Hence the resistance R is calculated as -

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

Now a graph is plotted taking potential V on X-axis and current I along Y- axis.

The graph will be a straight line for conductor and will be a curve depending on the type of insulator.

The slope of this graph will give the resistance of the material.

An insulator is a substance through which current flow is very low as its resistance is very high.Hence the slope of the V-I curve is very large.

A conductor is that substance through which current is flown easily as the resistance of conductor is very low.Hence the slope of V-I curve is smaller,

5 0

3 years ago

The moon has a mass of 1×10^22 kg, and the gravitational field strength at a distance r from the planet is 0.001 n/kg. what is t

CaHeK987 [17]

When I divided by 6, I got W(moon) = mg(moon) = 10 x 10/6 = 16.7N (the moon's gravity is only 1/6 that of Earth in this case). 100N is equal to W(Earth) = mg(Earth) = 10 x 10. On the moon and the Earth, m = 10 kg.

How can you determine the strength of Earth's gravitational field at the moon?

M = 6 10 24 K g is the mass of the earth, and g = - G M R 2 is the gravitational field of the earth at the point of the moon. The distance between the earth and moon is 84 10 8 meters. field of gravitation.

How do you determine the distance between Earth and the moon where the gravitational field is at its weakest?

You only discover points where the forces from all of the different local bodies are in balance, or equal to each other, because there is no place in the universe where the gravitational field intensity is zero. All bodies, not just the Earth and Moon, must be taken into account in the computation to arrive at these positions.

To know more about gravitational field strength at moon visit;

brainly.com/question/20909424

#SPJ4

4 0

2 years ago

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

Fittoniya [83]

Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be $0$ since the velocity of both astronauts was $0\!$ .

Therefore:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .