Answer:
a) 24.692 m/s
b) 19.4 m
Explanation:
To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:
We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:
a) The velocity (V2) is:
Substituting the known values we can get the velocity at the out:
Atmospheric pressure= 101000 Pa
Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3
b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:
We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:
Substituting the known values, the height (Z3) is:
Z3=Maximum Height=19.376=19.4 m
The formula we use
here is:
radial acceleration =
ω^2 * R <span>
110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>
<span>and since: ω = 2pi*f --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
<span>= 36,714.77 rpm </span></span>
A :-) F = mv^2 by t
Given - m = 10 kg
r = 10 m
v = 10 m/s
Solution -
F = mv^2 by t
F = 10 x 10^2 by 10
F = 10 x 100 by 10
( cut 10 and 10 because 10 x 1 = 10 )
F = 100 N
.:. The centripetal force is 100 N
Sorry but you tell me not to ask for help or answers but you want them yourself
Answer:
The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.
Explanation:
Given that,
Wavelength = 617 nm
Width of slit
Distance between the slit and screen L= 2.83 m
Third dark fringe m = 3
We need to calculate the distance on the screen from the center of the central bright fringe to the third dark fringe on either side
Using formula of distance
Put the value into the formula
Hence, The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.