Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu
1 answer:
Answer:
The value of gauge pressure at outlet = -38557.224 pascal
Explanation:
Apply Bernoulli' s Equation
+
+
=
+
+
--------------(1)
Where
= Gauge pressure at inlet = 3.70105 pascal
= velocity at inlet = 2.4
= Gauge pressure at outlet = we have to calculate
= velocity at outlet = 3.5
= 3.6 m
Put all the values in equation (1) we get,
⇒ +
=
+
+ 3.6
⇒ 0.294 = + 0.6244 + 3.6
⇒ = 0.294 - 0.6244 - 3.6
⇒ = - 3.9304
⇒ = - 38557.224 pascal
This is the value of gauge pressure at outlet.
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Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C