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sdas [7]
3 years ago
13

Which region of the Sun is typically about 6,000 C?

Physics
1 answer:
stich3 [128]3 years ago
4 0

the reign of the sun is very an important because the typically is about the size of 6,000 ceilometers

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DEFINE UNIFORM AND NON UNIFORM VELOCITY
Inessa05 [86]

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

7 0
3 years ago
Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show
ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = \frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}

We know that the inductance of the coil is given by the formula:

L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH

4 0
3 years ago
A 4 kg block is pulled with 4.1 N of force to the right. The block experiences 1.0 N of
Evgesh-ka [11]

Answer:

CHECK THIS..

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4 0
2 years ago
Which of the following would be the magnitude of the vector given a horizontal component of 30 and a
Scorpion4ik [409]

Answer:

50

Explanation:

Use the Pythagorean theorem to find the length of the diagonal, or the hypotenuse of an imaginary triangle. 30^2 + 40^2 = 2500, which is 50^2. So, the magnitude is 50.

Brainliest, please :)

7 0
1 year ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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