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3 years ago

A 12.0 khz, 16.0 v source connected to an inductor produces a 4.00 a current. what is the inductance?

1 answer:

4 0

Inductive reactance (Z) = ω L = 2Πf L = (2Π) (12,000) (L)

I = V / Z

4 A = 16v / (24,000Π L)

Multiply each side by (24,000 Π L):

96,000 Π L = 16v

Divide each side by (96,000 Π) :

L = 16 / 96,000Π = 5.305 x 10⁻⁵ Henry

L = 53.05 microHenry

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Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 $m/s^2$

Substituting

$520 = u*223\\ \\u = 2.33 m/s$

The constant velocity of car a = 2.33 m/s

b) We have $s = ut+\frac{1}{2} at^2$

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

$520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2$

Now we have v = u+at, where v is the final velocity

Substituting

v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 $m/s^2$

7 0

3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago

A good sign that cardiovascular exercise is occurring is that you _____.

grigory [225]

Are breathing hard. This is because cardiovascular exercise makes the heart beat faster which in turn creates a need for more air.

4 0

2 years ago

Read 2 more answers

Calculate the amount of force required to produce motion in a car of 1000kg with an acceleration of 5m/s square.

atroni [7]

Answer:

5000 N

Explanation:

F=ma

=1000*5

=5000N

4 0

3 years ago

Read 2 more answers

Help needed

MakcuM [25]

Answer: Atmosphere and geosphere.

Geo-sphere is the solid part of the earth. Hydrosphere is the water part. The living things on the earth make up the biosphere. The gases, water vapors etc make up the atmosphere. When a volcanic eruption adds carbon dioxide to the air, the carbon-dioxide is being added from geosphere to the atmosphere. Hence, there is interaction between atmosphere and geosphere.

6 0

3 years ago