Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components
Acceleration components
Time taken is 0.26088 seconds
The distance the beetle travels on the ground is 0.3677181864 m
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
Answer:
Explanation:
Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .
magnetic field in a torroid B = μ₀ n I , n is number of turns per unit length and I is current .
B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)
flux in the secondary coil
= B x area of face of coil x no of turns of secondary
= 4π x 10⁻⁷ x (1000 /2π x .16 ) .25 x 10⁻⁴ x 750
= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹
2343.75 x 10⁻⁸
= 23.43 x 0⁻⁶ H.
.
