Vector A and B are given as follows:

A rear window defroster consists of a long, flat wire bonded to the inside surface of the window. When current passes through th

GenaCL600 [577]

Answer:

$\rho=1.8\times 10^{-8}\Omega.m$

Explanation:

Given:

width of the wire, $w=1.8\ mm=1.8\times 10^{-3}\ m$

thickness of the flat wire, $d=0.12\ mm=1.2\times 10^{-4}$

length of the wire, $l=12\ m$

voltage across the wire, $V=12\ V$

current through the wire, $I=12\ A$

Now the net resistance of the wire:

using ohm's law

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

$R=\frac{12}{12}$

$R=1\ \Omega$

We have the relation between the resistivity and the resistance as:

$R=\rho.\frac{l}{a}$

where:

a = cross sectional area of the wire

$\rho =$ resistivity of the wire material

$1=\rho\times \frac{12}{1.8\times 10^{-3}\times 1.2\times 10^{-4}}$

$\rho=1.8\times 10^{-8}\Omega.m$

4 0

2 years ago

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[ $E_2 =\frac{2\times k \times q}{r^2}$

[ $r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$ ]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

3 0

2 years ago

Read 2 more answers

Energy that does not involve the large-scale motion or position of objects in a system is called:

Kryger [21]

I believe the answer is C.

4 0

2 years ago

You drop a small ball, and then a second small ball. When you drop the second ball, the distance between them is 3 cm. What stat

Alja [10]

Answer:

c) The distance between the balls increases.

Explanation:

If you drop the balls at the same time, regardless of their masses they accelerate equally, since they will be in free fall.

However, if you drop one of the balls earlier, then that ball will gain velocity, whereas the second ball has zero initial velocity. At the time the second ball is dropped, both balls have the same acceleration but different initial velocities.

According to the below kinematics equation:

$x = v_0t + \frac{1}{2}at^2$

The initial velocity of the first ball will make the difference, and the first ball will travel a greater distance than the second ball. Hence, their distance increases.

3 0

2 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$