Answer:
Explanation:
Given that
Length = L
At initial over hanging length = Xo
Lets take the length =X after time t
The velocity of length will become V
Now by energy conservation
So
We know that
At t= 0 ,X=Xo
So we can say that
So the length of cable after time t
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
This question involves the concepts of general gas equation and pressure.
The force exerted by the gas on one of the walls of the container is "74.08 KN".
First, we will use the general gas equation to find out the pressure of the gas:
where,
P = Pressure of the gas = ?
V = Volume of cube = (side length)³ = (10 cm)³ = (0.1 m)³ = 0.001 m³
n = no. of moles = 3 (since molecules equal to avogadro's number make up 1 mole)
R = general gas constant = 8.314 J/mol.K
T = Absolute Temperature = 24°C + 273 = 297 K
Therefore,
P = 7407.78 KPa
Now, the force on one wall can be given as follows:
where,
A = area of one wall = (side length)² = (0.1 m)² = 0.01 m²
Therefore,
<u>F = 74.08 KN</u>
<u></u>
Learn more about the general gas equation here: