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RoseWind [281]
3 years ago
5

How does activity on the Sun affect natural phenomena on Earth?

Physics
1 answer:
Nana76 [90]3 years ago
5 0

Answer and Explanation:

The Sun is the main source of energy on the earth if there will be no availability of Sun energy then life is impossible om the earth besides this the Sun warms our planet. The heating of ocean and atmosphere is mainly sue to Sun energy .Sun has also a great impact on the weather we can say that Sun is weather deciding on the earth our climate is totally dependent on the how much energy we got in form of radiation from earth.

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The distance between two consecutive nodesof a standing wave is 20.9cm.Thehandgen-erating the pulses moves up and down throughac
irina1246 [14]

Answer:

Velocity, v = 0.239 m/s

Explanation:

Given that,

The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m

The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.

For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

\dfrac{\lambda}{2}=0.209\ m\\\\\lambda=0.418\ m

Frequency is number of cycles per unit time.

f=\dfrac{2.57}{4.47}\\\\f=0.574\ Hz

Now we can find the velocity of the wave.

Velocity = frequency × wavelength

v = 0.574 × 0.418

v = 0.239 m/s

So, the velocity of the wave is 0.239 m/s.

4 0
2 years ago
In the diagram, the wavelength is shown by :
MAXImum [283]

Answer:

i think the anwer is C

Explanation:

6 0
3 years ago
How much mass energy could be obtained from the complete conversion of a 235 g hamburger?
Radda [10]
E = mc²
E = 0.235 kg · (3×10⁸ m/s)² = 0.235 · 9×10¹⁶  kg·m/s²
E = 2.115×10¹⁶ J
The answer is d) 2.12×10¹⁶ J
8 0
3 years ago
Why do elements need to be heated to emit color?
kirill115 [55]
Light carries away the energy. At a room temperature they are at their lowest energy
7 0
3 years ago
An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po
viktelen [127]

Answer:

1.232\times 10^{-5}\ T.

Explanation:

<u>Given:</u>

  • Current through the wire, passing through the origin, I_1 = 250\ A.
  • Current through the wire, passing through the y axis, r_y=1.8\ m., I_2 = 50\ A.

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to \mu_o times the net current threading the loop.

\oint \vec B \cdot d\vec l=\mu_o I.

In case of a circular loop, the directions of magnetic field and the line element d\vec l, both are along the tangent of the loop at that point, therefore, \vec B\cdot d\vec l = B\ dl.

\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\

\oint dl is the circumference of the Amperian loop = 2\pi r

Therefore,

B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_1 = 3.510\ m.

The magnetic field at the given point due to this wire is given by:

B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_2 = 5.310\ m.

The magnetic field at the given point due to this wire is given by:

B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at r_r=-3.510\ m is given by

B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.

6 0
3 years ago
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