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Kitty [74]
3 years ago
5

Radio communications were severely Distributed last week And the local newspaper says that solar activity is to blame . Which ty

pe of solar activity could the newspaper be talking about ?
Physics
1 answer:
Juliette [100K]3 years ago
8 0

Answer:

Solar Flares and Prominences

Explanation:

    The sun is source of light and heat energy and it supports life on earth because sun is the primary source of energy. But sun is also a star and is a glowing ball of hot gases mostly hydrogen and helium. There are so many activity that takes place on surface of sun. But the two activity solar flare and prominences affect the radio communication.

Solar Flare  are electromagnetic energy built up in the sun atmosphere .These energy travel with the speed of light and affect outer atmosphere of earth. These increased level of X rays and ultraviolet rays ions the layer of ionosphere which becomes superionised. thus causing destructive interference.

The radio waves travel in ionosphere and  frequently collide with electron lose energy and completely blackout.

Prominence is defined as gaseous cloud of calcium,hydrogen and other gases floating above sun chromosphere and emit violently particularly near sunspots sometimes travel through space and affect radio communication.

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What is the speed of the block/bullet system after the collision? express your answer in terms of vi, mw, and mb. hints?
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Read 2 more answers
Three cables AB, AC, and AD are supporting a 70-m tall tower. The magnitude of force in each cable is 2 kN. Find the total force
Mrac [35]

Explanation:

The coordinates of points are:

A (0, 70, 0), B (40, 0, 0), C (40, 0, 40) D (60, 0, 60).

The position vectors for corresponding cables are:

r_{AD} =(-60-0)i+(0-70)j+(-60-0)k\\r_{AD} =-60i-70j-60k\\r_{AC} =(-40-0)i+(0-70)j+(40-0)k\\r_{AC} =-40i-70j+40k

r_{AB} =(40-0)i+(0-70)j+(0-0)k\\r_{AB} =40i-70j+0k

the unit vectors for these positions are:

u_{AD} =r_{AD} /[r_{AD} ]=(-60/110)i-(70/110)j-(60/110)k\\=-0.5455i-0.6364j-0.5455k

u_{AC} =r_{AC} /[r_{AC} ]=-(40/90)i-(70/90)j+(40/90)k\\=-0.444i-0.778j+0.444k

u_{AB} =r_{AB}/[ r_{AB} ]=(40/80.6)j-(70/80.6)j+0k\\=0.4963i-0.8685j+0k

The factors are:

F_{AB} =[F_{AB} ]u_{AB} =0.9926i-1.737j+0k\\\\F_{AC}= [F_{AC]} u_{AC} =-0.8888i-1.5556j+0.8888k

F_{AD} =[F_{AD}] u_{AD} =-1.0910i-1.2728j-1.0910k

so The resultant force exerted on tower by cables are:

F_{R} =F_{AB}+ F_{AC}+ F_{AD} \\=-0.9875i-4.5648j-0.2020k\\magnitude=[F_{R} ]=4.674kN

6 0
3 years ago
Consider a U-tube whose arms are open to the atmosphere. Now water is poured into the U-tube from one arm, and light oil (r 5 79
EastWind [94]

Answer:

water height: 3cm

oil height: 12 cm

Explanation:

70 cm = 0.7 m

Suppose both arms have the same cross-section area S:

Let water density be \rho_w = 1000 kg/m^3

Let x be the height of water in the 2nd arm, then 4x is the heigh of oil in the 2nd arm.

For the system to be balanced, then the fluid mass of both arms must be the same

m_1 = m_2

V_1\rho_w = V_w\rho_w + V_o\rho_o

Sh_1\rho_w = Sx\rho_w + S4x\rho_o

Where V1 is the fluid volume  is the 1st arm, Vw is the water volume in the 2nd arm, Vo is the oil volume in the 2nd arm, h_1 = 0.7 m is the fluid height in the 1st arm, \rho_o = 5790kg/m^3 is the oil density.

We can divide both sides by S

h_1\rho_w = x\rho_w + 4x\rho_o

h_1\rho_w = x(\rho_w + 4\rho_o)

x = \frac{ h_1\rho_w }{(\rho_w + 4\rho_o } = \frac{0.7*1000}{1000 + 4*5790} = 0.03m = 3 cm

So the water height in the 2nd arm is 3 cm and the oil height is 3*4 = 12 cm

7 0
3 years ago
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