Answer:
An example in which liquid pressure phenomena can be used in daily life is in Water blasting
Explanation:
Water blasting refers application of pressurized water to remove materials from the surface of objects.
There are different varieties of water blasting, including;
Hydrocleaning; Cleaning enabled by the use of high pressure water
Hydrodemolition; Demolition or removal of concrete using pressurized water
Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.
The correct field line would be A.
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
Answer:
Hence the pressure is 
Explanation:
Given data
Q=1500 J system gains heat
ΔV=- 0.010 m^3 there is a decrease in volume
ΔU= 4500 J internal energy decrease
We know work done is
W= Q- ΔU
=1500-4500= -3000 J
The change in the volume at constant pressure is
ΔV= W/P
there fore P = W/ΔV= -3000/-0.01= 3×10^5
Hence the pressure is
Answer:4
Explanation:
Given
Frequency of beats is 3 beats per second
Frequency of first tuning fork is 
Beat frequency is difference in the frequency of tuning forks i.e.
either 
or 
so 
or
so there is not enough information given to decide.
