Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)
a) From the diagram, the horizontal component of the force is
Fₓ = 25 cos 20° = 23.49 N = 25 N
b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain
Net force = Fₓ - Frictional force
But frictional force is 0 N
Net force = Fₓ
Hope this Helps!!!
Answer:
Explanation:
Let T be the tension in the swing
At top point 
where v=velocity needed to complete circular path
r=distance between point of rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)
Th-resold velocity is given by 
To get the velocity at bottom conserve energy at Top and bottom
At top 
Energy at Bottom 
Comparing two as energy is conserved
Answer:
t = 224 s
Explanation:
given,
length of container = 15 cm = 0.15 m
diameter of spherical particle = 2.5 μm
mass of particle = 1.9 x 10⁻¹⁴ kg
viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s
time taken by the particle to stop = ?
radius of particle = 2.5/2 = 1.25 μm
volume of particle = 
=
Density =
Density =
ρ = 2322 kg/m³
terminal velocity
v_t = 6693 x 10⁻⁷ m/s
t = 224 s
Answer:
The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.
shdh
Answer:
(A) Work done will be 87.992 KJ
(B) Work done will be 167.4 KJ
Explanation:
We have given mass of methane m = 4.5 gram = 0.0045 kg
Volume occupies 
And volume is increased by 
so 
Temperature T = 310 K
Pressure is given as 200 Torr = 26664.5 Pa
(a) At constant pressure work done is given by
(b) At reversible process work done is given by 
We have given mass = 4.5 gram
Molar mass of methane = 16
So number of moles 
So work done 
