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2 years ago

A 1.5m wire carries a 3 A current when a potential difference of 74 V is applied. What is the resistance of the wire?

Physics

1 answer:

Allisa [31]2 years ago

4 0

Answer:

24.7 ohms

Explanation:

Given in the question,

electric current = 3 A

potential difference = 74 V

Using ohm's law

R = V/I

<em>R = resistance of wire</em>

<em>V = potential difference</em>

<em>I = electric current</em>

<em />

plug value in the formula

R = 74/3

R = 24.7 ohms

Therefore, the resistance of the wire is, 24.7 ohms

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1. During the Middle Ages, armies often attacked castles using large siege engines such as the counterweight trebuchet at left.

fredd [130]

Answer: Some challenges that I could see would be the walls, possibly a moat, tar, and the towers.

Explanation: The wall is obliviously a main problem, trying to get over it or through it is a difficult challenge. The moat (if it has one) means that there is more than likely only one way to get in or out. If it does have tar it means that the attackers are going to be put in a "sticky situation" And finally the towers, they have people at the top shooting arrows down at you, or throwing things at you.

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3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago

The value of acceleration due to gravity is less at the top of Mount Everest then that in the terai reason, why?

xxTIMURxx [149]

Answer:

The value of acceleration due to gravity is greater in terai than in mountain. In terai region the radius of earth is less as it lies close to the centre of the earth. Thus, the value of g is more in terai region.

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2 years ago

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

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3 years ago

How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.

hodyreva [135]

Answer:

Zero

Explanation:

W = F × s

F = 10 N,

t = 3min = 180sec

s = 0( no change in postion)

W = 10 ×0

W = 0

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3 years ago