<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>
Answer:
The total number of small cylinder = 7.
Explanation:
Lets take
Radius of the large cylinder = R
length = L
L = 10 R
The total area A = 2 π R² + π R L
The length of the small cylinder = l
The number of small cylinder = n
L = n l
The total area of small cylinders
A'=n (2 π R² + π R l)
As we know that emissive power given as
P = A ε σ T⁴
For large cylinder
P = A ε σ T⁴ -----------1
For small cylinders
P'=A' ε σ T⁴ ------2
From 1 and 2
Given that
P'= 2 P
A' ε σ T⁴ =2 A ε σ T⁴
A'=2 A (All others are constant)
n (2 π R² + π R l) =(2 2 π R² + π R L)
n (2 R² + R l) = (2 R² + R L)

L = 10 R


2 n +10 = 2 x 12
2 n +10 = 24
2 n = 24 -10
2 n = 14
n = 7
The total number of small cylinder = 7.
The kinetic energy equals the work, which is calculated by:
τ = F*d
τ = 300*2
τ = 600 J
τ = Ke = 600 J
<em>ANSWER</em>
<u>An increase in relative humidity</u>
<em><u>Could you mark me brainliest plz?</u></em>