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valentinak56 [21]
3 years ago
11

Find the resistance of a 0.03 m long copper wire with a radius of .005 m (Area of a circle: π r²).

Physics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

578feet ygti

bkyjuuhmjkungi9

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Consider the following equilibrium at 979 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium rea
GenaCL600 [577]

Answer:  I_2=  0.0050 M

I = 0.0155 M

Explanation:

Initial moles of  I_2 = 0.072 mole

Volume of container = 3.9 L

Initial concentration of I_2=\frac{moles}{volume}=\frac{0.072moles}{3.9L}=0.018M  

The given balanced equilibrium reaction is,

                 I_2(g)\rightleftharpoons 2I(g)

Initial conc.         0.018 M            0

At eqm. conc.    (0.018-x) M      (2x) M  

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[I]^2}{[I_2]}

K_c=\frac{(2x)^2}{0.2-x}

we are given :  K_c=1.60\times 10^{-3}

Now put all the given values in this expression, we get :

1.60\times 10^{-3}=\frac{(2x)^2}{(0.018-x)}

x=0.0025

So, the concentrations for the components at equilibrium are:

[I]=2\times x=2\times 0.0025=0.0050

[I_2]=0.018-x=0.018-0.0025=0.0155

Hence, concentrations of I_2 and I are 0.0050 M ad 0.0155 M respectively.

4 0
4 years ago
A train is travelling at 15m/s. It accelerates at 3m/s² for 20 seconds. How far does the train travel in this time?
Fittoniya [83]

Answer:

Explanation:

The answer is 900m

4 0
3 years ago
Which of the circuit diagrams shown in Figure 21-1A is a parallel circuit?
Thepotemich [5.8K]
I and II only it’s has multiple paths for the electricity to flow
4 0
3 years ago
Plz answer fast the question
alexgriva [62]

Answer:

Angle of incidence = 20°

Angle of reflection = 20°

Explanation:

Applying,

The first Law of Refraction: The incident ray, the reflected ray and the normal at the point of incidence all lies in the plane.

From the diagram,

Angle of incidence = 90-70

Angle of incidence = 20°

From the law of reflection,

Angle of incidence = Angle of reflection

Therefore,

Angle of reflection = 20°

3 0
3 years ago
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
Gnom [1K]

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

The net Force F = ma

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

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
3 years ago
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