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Ainat [17]
3 years ago
7

Sunspots are located

Physics
2 answers:
MA_775_DIABLO [31]3 years ago
6 0
Lines of longitude and latitude . Surprisingly, almost all sunspots are found in two bands on the Sun, just north and south of the equator. At the start of the cycle, the sunspots appear at latitudes of about 30 degrees, both north and south<span> of the equator.</span>
LUCKY_DIMON [66]3 years ago
5 0

Answer:

Sunspots are located on the Sun's photosphere.

Explanation:

:)

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What is the wave speed of a 1,000 Hz wave with a wavelength of 0.04 m?
dalvyx [7]

Wave speed = 40 m/s

Explanation:

Wave speed = frequency x wavelength

Wave speed = 1000 * 0.04

Wave speed = 40 m/s

6 0
3 years ago
A physicist comes across an open container which is filled with two liquids. Since the two liquids have different density, there
Dominik [7]

Answer:

496.57492 kg/m³

Explanation:

P_a = Atmospheric pressure = 101300 Pa

\rho_w = Density of water = 1000 kg/m^3

h_w = Height of water = 21.8 cm

h_f = Height of fluid = 30 cm

g = Acceleration due to gravity = 9.81 m/s²

\rho_f = Density of the unknown fluid

Absolute pressure at the bottom

P_{abs}=P_a+\rho_wgh_w+\rho_fgh_f\\\Rightarrow \rho_f=\frac{P_{abs}-P_a-\rho_wgh_w}{gh_f}\\\Rightarrow \rho_f=\frac{104900-101300-1000\times 9.81\times 0.218}{9.81\times 0.3}\\\Rightarrow \rho_f=496.57492\ kg/m^3

The density of the unknown fluid is 496.57492 kg/m³

6 0
3 years ago
Band of which of these colours are not seen in a spectrum
Ede4ka [16]
The only colours that are in the spectrum are red, orange, yellow, green, blue, indigo and violet. hope this helps!
7 0
3 years ago
How much work is done by the gravitational force when a 265-kg pile driver falls 2.80m?​
sp2606 [1]

Answer:

i dont know why are u asking me

Explanation:

6 0
3 years ago
A solenoidal coil with 21 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 25.0 cm long
77julia77 [94]

A) We need to calculate the Magnetic flux of a Solenoid,

\Phi = BA

Where B is the magnetic field and A the Area.

B=\frac{\mu_0 N_2 I}{l}

Here \mu_0 is the permeability constant, I the current and N number of turns.

Replacing we have,

\Phi = \frac{\mu_0 N_2 I}{l}(\pi r^2)

\Phi = \frac{(4\pi*10^{-7})(350)(0.12)}{25*10^{-2}} (\pi*\frac{2.5*10^{-2}}{2}^2)

\Phi = 1.036*10^{-7}Wb

B) We calculate the mutual inductance, so

M=\frac{N_1 \Phi}{i}

M=\frac{21*1.036*10^{-7}}{0.12}

M=1.813*10^{-5}H

c) We calculate the emf

\epsilon = -M \frac{dI}{dt}

\epsilon = -1.813*10^{-5}*1700

\epsilon= -0.030V

3 0
3 years ago
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