Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
Explanation:
From the question we are told that
The time constant 
The potential across the capacitor can be mathematically represented as
Where 
is the voltage of the capacitor when it is fully charged
So at
Generally energy stored in a capacitor is mathematically represented as
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at
The energy stored can be evaluated at as
Hence the fraction of the energy stored in an initially uncharged capacitor is
Given:
u(initial velocity)=0
v(final velocity)= 10 m/s
t= 4 sec
Now we know that
v= u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration measured in m/s^2
t is the time measured in sec
10=0+ax4
a=10/4
a=2.5 m/s^2
Answer:
is there an equasion it gives you?
Explanation:
need a little more info.
Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
9.8........................
