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Alinara [238K]
2 years ago
8

Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr

oximate voltage drop will he experience? A. 1.8 V B. 2.8 V C. 3.5 V D. 5.9 V
Physics
1 answer:
castortr0y [4]2 years ago
8 0

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

R = \frac{2.525}{1000}\times 50

R = 0.126 ohm

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

V = iR

V = 11 \times 0.126

V = 1.4 Volts

so most appropriate answer in given options is

A. 1.8 Volts

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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
2 years ago
A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
evablogger [386]

Answer:

\text{heat loss} = 24864.05 \  W/m^2

Explanation:

If

  • T_1, T_2 are temperatures of gasses and liquid in Kelvins,
  • t_1 and t_2 are thicknesses of gas layer and steel slab in meters,
  • h_1, h_2 are convection coefficients gas and liquid in W/m^2 \cdot K,
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then: part(a):

\text{heat loss } =  \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

using known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Using the rate equation :

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Similarly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature distribution is shown in the attached image

3 0
2 years ago
A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000
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We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). <span>

We want a ratio of 3000000 to 1, or 3 million to 1. 

</span>

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<span>v = 3 x 10^8 m/s</span></span></span>

8 0
2 years ago
A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
pochemuha
10,000 units of momentum.
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Jane puts some water into an electric kettle and then she connects it to the power source. She observes that after some time the
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C) electrical energy is transformed into heat energy
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2 years ago
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