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Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the tex

OlgaM077 [116]

Answer:

L = 0.109 H

Explanation:

Given that,

Number of loops in the solenoid, N = 1500

Radius of the wire, r = 4 cm = 0.04 m

Length of the rod, l = 13 cm = 0.13 m

To find,

Self inductance in the solenoid

Solution,

The expression for the self inductance of the solenoid is given by :

$L=\dfrac{\mu_o N^2 A}{l}$

$L=\dfrac{4\pi \times 10^{-7}\times (1500)^2\times \pi (0.04)^2}{0.13}$

L = 0.109 H

So, the self inductance of the solenoid is 0.109 henries.

3 0

3 years ago

A microwave oven has 1200 W. If it receives 120 V of potential difference, what is the current in the microwave?

8_murik_8 [283]

Answer:

10amps

Explanation:

Given data

Power =1200W

Voltage =120V

Required

The current I

From

P=IV

I= P/V

Substitute

I=1200/120

I= 10amps

Hence the current is 10amps

7 0

2 years ago

A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com

blagie [28]

Answer:

4500.5 nutritional calories per gram

Explanation:

Heat lost by the new candy = heat gained by the bomb calorimeter.

Heat gained by the bomb calorimeter = c×ΔT

where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K

ΔT = change in temperature = 2.69°C = 2.69 K.

Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J

Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

8 0

3 years ago

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

elena55 [62]

Answer:

$3.2\times 10^{-7}\ m$ or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = $9.38\times 10^{14}\ Hz$

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = $3\times 10^8\ m/s$

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

$v=f\lambda$

Now, expressing the above equation in terms of wavelength 'λ', we have:

$\lambda=\frac{v}{f}$

Now, plug in the given values and solve for 'λ'. This gives,

$\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m$

Therefore, the wavelength of the radiations absorbed by the ozone is nearly $3.2\times 10^{-7}\ m$ or 0.32 μm.

7 0

3 years ago

Why are there so few working wave power stations?

Bond [772]

Well mostly because it's for oceans and lakes that have waves. so if you don't live near any of these, you wouldn't be getting power.

3 0

3 years ago

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