Answer:
At point A, the cart has high potential energy. At point b, the cart is pulled down by gravity. At point c, the cart gains its highest kinetic energy. At point d, the cart returns back to the same state but with lower potential energy.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
<u>FOR BULB 1:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
<u>FOR BULB 2:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
<u>A₁/A₂ = 0.44</u>
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
