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fenix001 [56]
3 years ago
13

water flows in a pipe of a terminal of A1=0.02 m2 with a speed of V1=2 m/s. At the second end of the pipe A2=0.04 m2, the speed

of flow is V2=_______m/s
Physics
1 answer:
weeeeeb [17]3 years ago
8 0
A1*V1=A2*V2
0.02*2=0.04*V2
V2=1 m/s
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Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
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Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

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So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

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a = 640 m/s²

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