Answer:
v = 6.315 cm
Explanation:
given,
R₁ = 4 cm = 0.04 m
R₂ = 15 cm = 0.15 m
n =1.5
v = 0.06315 m
v = 6.315 cm
hence, the distance of the image from the back surface is v = 6.315 cm
Ella makes this table to organize her notes on whether atoms gain or lose energy during the changes of state.
2 answers:
You might be interested in
Answer:
29,7 m
Explanation:
We need to devide the problem in two parts:
A) Energy
B) MRUV
<u>Energy:</u>
Since no friction between pint (1) and (2), then the energy conservatets:
Energy = constant ----> Ek(cinética) + Ep(potencial) = constant
Ek1 + Ep1 = Ek2 + Ep2
Ek1 = 0 ; because V1 is zero (the ball is "dropped")
Ep1 = m*g*H1
Ep2= m*g*H2
Then:
Ek2 = m*g*(H1-H2)
By definition of cinetic energy:
m*(V2)²/2 = m*g*(H1-H2) --->
Replaced values: V2 = 14,0 m/s
<u>MRUV:</u>
The decomposition of the velocity (V2), gives a for the horizontal component:
V2x = V2*cos(α)
Then the traveled distance is:
X = V2*cos(α)*t.... but what time?
The time what takes the ball hit the ground.
Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²
In the vertical axis:
Y3 = 0 ; Y2 = H2 = 2 m
Reeplacing:
-2 = 14*t + (1/2)*(-9,81)*t²
solving the ecuation, the only positive solution is:
t = 2,99 sec ≈ 3 sec
Then, for the distance:
X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m
