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jarptica [38.1K]
4 years ago
13

Some examples of solar ejections

Physics
1 answer:
mylen [45]4 years ago
8 0
Coronal Mass ejections or Solar ejections are activities on the surface of the sun termed to the reaction of the gas composition of the sun's surface attributing to explosion of these gases. One example most commonly known as Solar Flares, and another example is termed as erupting prominence.
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Which occurs when a wave is reflected
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Reflection. It occurs when a wave bounces from the surface of an obstacle
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What test may be used to determine if a stain is blood or not
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As of I believe</span>
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3 years ago
A car of mass 650 kg slows from a speed of 20.0 m/s to a speed of 10.0 m/s. How much kinetic energy has it lost. How much force
Natalija [7]
Formulas you need for this problem:
F= mass•acceleration
KE= mass•velocity^/2
Acceleration= final velocity-intial velocity/time
time= distance/speed

t= 97.5\10= 9.75 seconds
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These are the answers :)
F=650•-1.03= -669.5N
KE= 650•100/2= 32,500J or 32.5KJ
7 0
4 years ago
Energy can be transferred as heat in many ways. Which statement is the best description of energy transfer by convection?
vaieri [72.5K]

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3 0
3 years ago
How much sweat (in ml ) would you have to evaporate per hour to remove the same amount of heat a 150 w light bulb produces in an
allsm [11]
<span>The first thing to do here is to figure out how many joules of energy would be produced by your lightbulb in one hour. As you can see, the conversion factor to use here is #"1 W" = "1 J"/"1 s"# This is basically a reminder of the fact that the watt, a unit of power, is defined as an energy of one joule delivered in one second. Since one hour is known to have #60# minutes, i.e. #60 xx "60 s"#, you will have #60 xx 60 color(red)(cancel(color(black)("s"))) * "140 J"/(1color(red)(cancel(color(black)("s")))) = "504000 J" = "504 kJ"# Your next step here will be to use the enthalpy of vaporization of water to figure out how many grams of water would be evaporated by that much heat. #DeltaH_"vap" = "44.66 kJ mol"^(-1)# Convert this to kilojoules per gram by using water's molar mass #44.66 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/("18.015 g") = "2.479 kJ g"^(-1)# This means that the mass of water that can be evaporated by #"504 kJ"# of heat will be #504color(red)(cancel(color(black)("kJ"))) * ("1 g H"_2"O")/(2.497color(red)(cancel(color(black)("kJ")))) = "201.8 g"# Now, assuming that the sweat is pure water, you can approximate its density to be equal to #"1.0 g mL"^(-1)#. This means that the volume of water that can be evaporated will be #201.8 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.0color(red)(cancel(color(black)("g")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(2.0 * 10^2 "mL")color(white)(a/a)|)))# The answer is rounded to two</span>
7 0
3 years ago
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