Answer:
An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
In a series circuit, all of the components are connected in the same 'loop' and the current only has one direction/path it can flow through.
In the first three options, the current has multiple paths it can go through. So these three circuits are parallel and not series.
In the last option, the current only has one path where it can flow through, so that circuit is in series.
So Circuit <u>D </u>is a series circuit.
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Answer
Circuit D
Wind speed and air temperature are used to calculate a windchill factor.
<u>Explanation:</u>
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Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.
The air temperature value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.
There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.
To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have





Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s