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pantera1 [17]
3 years ago
11

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie

s with L3, a 20 mH coil. The entire combination is across an ac source of 60 Vrms at 300 kHz. The total rms current is
Physics
1 answer:
Bond [772]3 years ago
7 0

Answer:

Explanation:

For parallel inductors ,

\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}

\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}

L_R=16.67 mH.

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ωL_{total} where ω is angular frequency

= 2πfL_{total}

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

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The density of the substance is<u> 10.5 g/cm³.</u>

The jewelry is made out of <u>Silver.</u>

Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume <em>V</em> of the jewelry is given by,

V=V_2 -V_1

Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

V=V_2 -V_1\\ =61.2 ml -48.6 ml\\ =12.6 ml

calculate the density ρ of the jewelry using the expression,

\rho =\frac{m}{V}

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml

Since 1 ml=1 cm^3,

The density of the jewelry is <u> 10.5 g/cm³.</u>

From standard tables, it can be seen that the substance used to make the jewelry is <u>silver</u><em><u>, </u></em>which has a density 10.5 g/cm³.



4 0
3 years ago
HELPP ME IN PHYSICS +15 POINTS!! <br> But a right answer not links or I’ll report. -_-
frozen [14]

The answer shud be 31 . 14m

8 0
3 years ago
A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes
mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

d_1 = v* t

d_1 = 20 * 0.8

d_1 = 16 m

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2*(-7)*d_2

d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

3 0
3 years ago
A block of ice (m = 9 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of t
jonny [76]

Answer:

a) An expression for the amount of energy, E_m, needed to melt the ice into water.

(E_m) = (m × Lf)

b) An expression for the total amount of energy, E_tot, to melt the ice and then bring the water to T2

(Total heat) = (m × Lf) + mc (T2 - T1)

c) 3,646,458 J = 3646.46 kJ

Explanation:

a) When a pure body changes its phase at meltimgbor boiling point, it does so at a constant temperature. When a pure body melts, the amount of heat responsible for this change is just given by a product od the mass of the body and the body's heat of fusion.

(E_m) = (m × Lf)

b) The Heat required to raise the temperature of a body from one temperature to another is given by the product of the mass of the body, its specific heat capacity and the temperature difference between the final point and the starting point.

(E_2) = mcΔT = mc (T2 - T1)

Total heat required to melt the ice at T1 = 0 and raise the temperature of the resulting water to T2 is then a sum of (E_m) + (E_2)

(Total heat) = (m × Lf) + mc (T2 - T1)

c) What is the energy in Joules?

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m = mass of ice = resulting mass of water = 9 kg

Lf = latent heat of fusion = 334000 J/kg

c = Specific heat capacity of water = 4186 J/kg.K

T2 = final temperature of the water = 17°C

T1 = Initial temperature of the water = 0°C

Note that the units of temperature difference is the same for K and °C

(Total heat) = (m × Lf) + mc (T2 - T1)

Q = (9 × 334000) + [9 × 4186 × (17 - 0)]

Q = 3,006,000 + 640,458 = 3,646,458 J = 3646.46 kJ

Hope this Helps!!!

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A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
blondinia [14]
The answer is D. Hope this helps you!
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