Explanation:
As the earth travels around the sun in the elliptical orbit it must also be know that the axis of the earth is tilted as well.
- So when the earth is at the farthest point and the tilt of the earth's axis is towards the sun in that case the sun rays always incident on the surface of the earth near the poles and hence there is sunlight for the 24 hours of the day. But the intensity of these rays is very low because of the their slanted angle of incident. In other words the same sun rays cover a larger area and the luminous intensity is reduced.
- When the earth is near to the sun we have an increased average temperature of the day during that phase giving us an experience of summer season and vice-versa is the condition in winter seasons. The tilt of the earths axis is responsible for variation in extremities of the seasons with respect to the geographical location.
D=vt use this equation to get the depth of the skull
Kinetic energy = (1/2) (mass) (speed²).
A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.
A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.
They're both correct.
Answer:
Limewater can be used to detect carbon dioxide. If carbon dioxide is bubbled through limewater then it turns from clear to cloudy/milky in colour. This is why limewater used in a simple respirometer can show that more carbon dioxide is present in exhaled air compared to inhaled air.
Explanation:
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
