I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
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For question 1 if you have your acceleration equal to -9.8m/s^2 then you can find how fast it is traveling when it hits the water. You would use the equation Vf=Vi+a(t). You are looking for your Vf. Vi is 0m/s. a is -9.8m/s^2. t is 3.87s. If you now plug in what you have and solve you would end up with Vf=-37.926. For how far did it drop use the equation, d=Vi(t)+1/2(a)(t)^2. plug in the Vi, a, and t and solve. Then you are left with the distance vertically being 73.38681m (don't forget sig figs). (When you solve for d it ends up being negative, but because you can't ever go backwards in distance it has to be positive)
Answer:
0.8 m/s²
Explanation:
Convert km/h to m/s:
20 km/h × (1000 m / km) × (1 h / 3600 s) = 5.56 m/s
50 km/h × (1000 m / km) × (1 h / 3600 s) = 13.89 m/s
Acceleration is change in velocity over change in time:
a = Δv / Δt
a = (13.89 m/s − 5.56 m/s) / 10 s
a = 0.83 m/s²
Rounded to one significant figure, the car accelerates at 0.8 m/s².
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.