Answer:
The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]
Explanation:
We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.
For this case, we don't use the elevation difference and therefore those terms can be cancelled.
When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.
Answer:
Explanation:
Hi!
The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.
The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:
Then the electric field at the point of interest is estimated as:
Answer:
Explanation:
according to third equation of motion
2as=vf²-vi²
vf²=2as+vi²
vf=√2as+vi²
vf=√2as+vi
vf=√2*2*4+3
vf=√16+3
vf=4+3=7
so final velocity is 7 m/s
Answer:
Explanation:
Stored energy in spring = 1/2 k x² , k is spring constant , x is compression.
= 1/2 x 8 x (5.7 x 10⁻²)²
= 129.96 x 10⁻⁴ J
Energy lost due to friction = force x distance
= .035 x .17
= .00595 J
Energy used in providing kinetic energy to projectile.
129.96 x 10⁻⁴ - .00595
.012996 - .00595
= .007046 J
So
1/2 m v² = .007046
v² = .007046 x 2 / .0059
= 2.3885
v = 1.545 m /s
P1v1/t1 = p2v2/t2
p1=475, v1=4, t1=290
v2=6.5, t2=277
solve for p2 in kpa
