Factors that influence fluid pressure are depth and “ “ of the fluid.

Physics

1 answer:

Ivenika [448]4 years ago

8 0

Explanation:

Two factors influence the pressure of fluids. They are the depth of the fluid and its density. A fluid exerts more pressure at greater depths. Deeper in a fluid, all of the fluid above it results in more weight pressing down.

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Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of

lions [1.4K]

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i² = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

= 0.2 A approx .

2 i = .4 A Ans .

6 0

4 years ago

Bro why cant I post this T-T

Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let $\alpha$ be the angular acceleration of the end of the rod

Taking torque about the link, we have:

$\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)$

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

$\tau = I_{rod}\ \alpha......(ii)$

From equations (i) and (ii) we have:

$mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }$