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Marina CMI [18]
3 years ago
12

The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid

in the rest of the pipe
Physics
1 answer:
timofeeve [1]3 years ago
4 0

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

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B It is moving in one dimension

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Three-fourths of the elements on the
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A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge
Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

6 0
3 years ago
A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

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What's the name of the compound Kl?
Nat2105 [25]

Answer: Potassium iodide

Explanation: their you go

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