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Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to

Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x ( $\frac{Rc}{Rc + Ri}$ )

where

Rc = resistance of copper = $\frac{ρl}{a}$ (l = length , a = area, ρ = resistivity of copper)

Ri = resistance of iron = $\frac{ρ₀l}{a}$ (l = length , a = area, ρ₀ = resistivity of copper)

Vc = V x ( $\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}}$ )

Vc = V x ( $\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})}$ )

Vc = V x ( $\frac{ρ}{ρ + ρ₀}$ )

where

ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter

ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x ( $\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}}$ )

Vc = 2.41 v

5 0

3 years ago

What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?

arlik [135]

Answer: 0.25 m/s

Explanation: Speed = wavelengt · frequency

v = λf and frequency is 1/period f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0

2 years ago

Sometimes you can see a faint reflection in the surface of a shiny plate or cup.Why ?

bazaltina [42]

Answer:

The degree of reflection whether faint or bright you see on the surface of an object is an indication that light particles had hit the surface. Since light is a wave and as part of its characteristics can get reflected. However, the amount of light reflected by a surface is dependent on the smoothness of the surface which can be shiny or dull, it can also be dependent on the nature of the surface which can be glass, water, and so on. So, from the question, you can see a faint reflection on the surface of a shiny plate or cup because of the smoothness of the surface which reflects the lights that hit it from a particular direction at the same angle.

4 0

2 years ago

A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui

GalinKa [24]

Answer:

$n_l = 1.97$

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

$\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

focal length of lens in liquid is

$\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}$

rearrange fro $n_l$

$n_l = \frac{n_g f_l}{f_l+f(n_g-1)}$

$n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}$

$n_l = 1.97$

7 0

3 years ago

Suppose a clay model of a koala bear has a mass of 0.205 kg and slides on ice at a speed of 0.720 m/s. It runs into another clay

Westkost [7]

Answer:

Final velocity, v = 0.28 m/s

Explanation:

Given that,

Mass of the model, $m_1=0.205\ kg$

Speed of the model, $u_1=0.72\ m/s$

Mass of another model, $m_2=0.32\ kg$

Initial speed of another model, $u_2=0$

To find,

Final velocity

Solution,

Let V is the final velocity. As both being soft clay, they naturally stick together. It is a case of inelastic collision. Using the conservation of linear momentum to find it as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{0.205\times 0.72+0.32\times 0}{(0.205+0.32)}$

V = 0.28 m/s

So, their final velocity is 0.28 m/s. Hence, this is the required solution.

6 0

3 years ago