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vova2212 [387]
3 years ago
11

What are solar flares

Physics
2 answers:
Bad White [126]3 years ago
8 0
It's a brief of intense high energy radiation from the sun's surface. When it happens it disrupts the earth's communication and power line transmissions
crimeas [40]3 years ago
4 0

Solar flares are vast clouds of electrically charged subatomic particles that are continually blasted away from the Sun.

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4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strik
nikklg [1K]

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

5 0
3 years ago
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Koi to inbox kr do......​
rosijanka [135]

Answer:

<h2>Tum karo yaar </h2>

<h2>INBOX.....</h2>

<h2>MARK AS A BRAINLIEST</h2>

<h2>PLEASE☆☆☆</h2>

4 0
3 years ago
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How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/
Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J

Therefore, amount of heat lost in this process is 16.736 J.

4 0
3 years ago
Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s
g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

v_{x} =  v cos(∅)

v_{y} =  v sin(∅)

We know that for a projectile motion,

t =\frac{2vsin(∅)}{g}

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = v_{x}×t = vcos(∅)×\frac{2vsin(∅)}{g} = \frac{v^{2}sin(2∅) }{g}.

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = \frac{v^{2}sin(2α) }{g} = \frac{v^{2}sin(2[90-β]) }{g} =\frac{v^{2}sin(180-2β) }{g} = \frac{v^{2}sin(2β) }{g} .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

     β = 90-25 = 65°

∴ The required angle is 65°.

5 0
3 years ago
Read 2 more answers
A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th
12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, L=10\ cm

The Elecric flux through one of the side of the cube is, \phi =-1 kNm^2/C.

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by \epsilon_0

Since Flux is a scalar quantity. It can added to get total flux through the surface.

\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C

So the the charge at the centre is calculated.

5 0
3 years ago
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