Which is not the most obvious difference between the terrestrial and jovian planets?

Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in

Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$

3 0

3 years ago

A gas has a volume of 6 L, a temperature of 70 degrees C, and a pressure of 2 atm. If the gas is compressed to a volume of 4 L,

Lelu [443]

I would say check the back of the book

4 0

3 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

4 years ago

If the pendulum took longer to complete one oscillation, how would the graph change?

pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.

If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.

If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

line would be less.

If it was a graph of the pendulum's mass, weight, temperature, cost, etc.,

then the graph would be a horizontal line, and nothing that might change

the period of oscillation would have any effect on the graph.

7 0

3 years ago

Read 2 more answers

Please help meh need answers

boyakko [2]

Netiquette is a list of all acceptable acronyms when texting. Netiquette refers to etiquette, or protocols and norms for communication, on the Internet.

4 0

3 years ago