Explanation:
<h2>
<em>The </em><em>S. </em><em>I. </em><em> </em><em>unit </em><em>of </em><em>momentum </em><em>is </em><em>Kg. </em><em>m/</em><em>s</em><em>e</em><em>c</em></h2>
<em>hope </em><em>it </em><em>helps </em><em>you </em>
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
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:
During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy
mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .
mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²
gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²
g = 1 / 4 x ω² r + 1 / 2 x ω² r
g = 3 x ω² r/ 4
ω² = 4g /3 r
= 4 x 9.8 / 3 x .25
= 52.26
ω = 7.23 rad / s .
Answer:
The correct option is;
Force of Friction
Explanation:
As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path
The motion equation is
v = ωr and we have the centripetal acceleration given by
α = ω²r and therefore centripetal force is then
m×α = m × ω²r = m × v²/r
The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.
