Which of the following values are in the domain of the function graphed

A wire with a resistance of 10 ohm is stretched so that its new length is four times its original length. find the resistance of

amm1812

Let original length be L. The new length is therefore 4L.

Let original cross sectional surface area of the wire be equal to πr^2.

This means original volume was L x πr^2 = Lπr^2

The volume is the same but the length is different so 4L x new surface area must be equal to Lπr^2. Let new surface area be equal to Y.

4L x Y = Lπr^2
=> Y = (πr^2 )/ 4

Using the resistivity formula,
R = pL/A. p which is resistivity is a constant so it stays the same

But this time, instead of L we have 4L and instead of πr^2 we have (πr^2)/4.

so the new resistance
= (4Lp)/ {(πr^2)/4}
= 16 (pL)/(πr^2)
= 16 (pL)/A. because πr^2 is A
since pL/A is equal to R from the formula, this is equal to
16 R.

R was 10 ohms
therefore new resistance is 16 x 10 = 160 ohms

6 0

4 years ago

If the current in each wire is the same, which wire produces the strongest magnetic field?

lions [1.4K]

Answer: option 4: A wire that is 2-mm thick and coiled.

Explanation:

The current in each wire is same. The magnetic field due to a current carrying wire increases if the wire is coiled with the more number of turns. A thick wire would cause low resistance to the current. Hence, a 2-mm thick wire which is coiled would produce the strongest magnetic field.

3 0

4 years ago

Read 2 more answers

Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30

vekshin1

66.7 Watts

Explanation:

Let $R_{1}=1.0$ ohms, $R_{2}=3.0$ ohms and $R_{3}=6.0\:$ ohms. Since $R_{2}$ and $R_{3}$ are in parallel, their combined resistance $R_{23}$ is given by

$\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

or

$R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms$

The total current flowing through the circuit <em>I</em> is given

$I=\dfrac{V_{s}}{R_{Total}}$

where

$R_{Total}=R_{1}+R_{23}= 3.0\:ohms$

Therefore, the total current through the circuit is

$I=\dfrac{30\:V}{3.0\:ohms}=10\:A$

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor $V_{1}$ :

$V_{1}=(10\:A)(1.0\:ohms)=10\:V$

This means that voltage drop across the 6-ohm resistor $V_{3}$ is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by

$P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W$

4 0

3 years ago

Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitu

ANEK [815]

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

$\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big)$ )

$\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)$

Similarly;

$\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)$

However; the relative difference is: $\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}$

$\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}$

$= 0.2828 \\ \\ \mathbf{\simeq 29\%}$

3 0

3 years ago

What is the half-life of an isotope if after 30 days you have 31.25 g remaining from a 250 g beginning sample size?

Dima020 [189]

Answer:The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ.

Explanation:

4 0

2 years ago