Answer:
166.67 N
Explanation:
Applying Pascal's principle,
Presure in the smaller piston(P') = Pressure in the bigger piston(P)
But,
Pressure = Force/Area
Pressure in the smaller piston(P') = Force applied to the smaller piston(F')/Area of the smaller piston(A')
Pressure in the bigger piston(P) = Force applied to the bigger piston(F)/Area of the bigger piston(A)
F'/A' = F/A.................. Equation 1
Make F the subject of the equation
F = F'A/A'.............. Equation 2
From the question,
Given: F' = 20 N, A = 120 cm², A' = 10 dm² = (10×100) = 1000 cm²
Substitute these values into equation 2
F = 20(1000)/120
F = 166.67 N
Fg = ma
Fg = (0.1143 kg) (9.81 N/kg)
Fg = 1.12 N
Answer:
a) x = 1.5 *10⁻⁴cos(524πt) m
b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s
a = -1.5 *10⁻⁴(524π)²cos(524πt) m/s²
c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm
x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm
Explanation:
x = Acos(ωt)
ω = 2πf = 2π(262) = 524π rad/s
x = 1.5 *10⁻⁴cos(524πt)
v = y' = -Aωsin(ωt)
v = -1.5 *10⁻⁴(524π)sin(524πt)
a = v' = -Aω²cos(ωt)
a = -1.5 *10⁻⁴(524π)²cos(524πt)
not sure about the last part as time is generally not given in mm
I will show at 1 second and at 0.001 s to try to cover bases
x(1) = 1.5 *10⁻⁴cos(524π(1))
x(1) = 1.5 *10⁻⁴cos(524π)
x(1) = 1.5 *10⁻⁴(1)
x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm
x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))
x(0.001) = 1.5 *10⁻⁴cos(0.524π)
x(0.001) = 1.5 *10⁻⁴(-0.0753268)
x(0.001) = -1.129902...*10⁻⁵ m
x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm
Both momentum and kinetic energy are conserved in elastic collisions (assuming that this collision is perfectly elastic, meaning no net loss in kinetic energy)
To find the final velocity of the second ball you have to use the conversation of momentum:
*i is initial and f is final*
Δpi = Δpf
So the mass and velocity of each of the balls before and after the collision must be equal so
Let one ball be ball 1 and the other be ball 2
m₁ = 0.17kg
v₁i = 0.75 m/s
m₂ = 0.17kg
v₂i = 0.65 m/s
v₂f = 0.5
m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f
Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so....
m(v₁i + v₂i) = m(v₁f + v₂f)
The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so....
v₁i + v₂i = v₁f + v₂f
Now we want the final velocity of the second ball so we need v₂f
so...
(v₁i + v₂i) - v₁f = v₂f
Plug in the numbers now:
(0.75 + 0.65) - 0.5 = v₂f
v₂f = 0.9 m/s
