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3 years ago

What information should be posted on or near a dishwasher?

Physics

1 answer:

Murljashka [212]3 years ago

7 0

A.. in case of any problems that may occur you would know what company to call

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An LC circuit consists of a 3.14 mH inductor and a 5.08 µF capacitor. (a) Find its impedance at 55.7 Hz. 563.57 Correct: Your an

ANEK [815]

Answer:

$z=561.7$

$z=214.1$

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

f = frequency = 55.7 Hz

Impedance is given as

$z=\frac{1}{2\pi fC} - 2\pi fL$

$z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)$

$z=561.7$

f = frequency = 11000 Hz

Impedance is given as

$z= - \frac{1}{2\pi fC} + 2\pi fL$

$z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)$

$z=214.1$

4 0

3 years ago

Which of the following is the SI unit for volume

Katarina [22]

Meters for mass kilograms for volume cubic meters for density kilograms per cubic meter

3 0

3 years ago

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m

harina [27]

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

6 0

3 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

3 years ago

Which of the following is an example of a force?

strojnjashka [21]

The answer would be d!

6 0

3 years ago