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3 years ago

How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?

Physics

1 answer:

Marrrta [24]3 years ago

8 0

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by $V = R I$

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

$115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes$

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Answer:

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2 years ago

A large beaker of water is placed on a scale, and the scale reads 40 N. A block of wood with a weight of 10 N is then placed in

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The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It

katrin2010 [14]

Answer:

Part a)

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

$KE_{tot} = 2.68 \times 10^{33} J$

Explanation:

Time period of Earth about Sun is 1 Year

so it is

$T = 1 year = 3.15 \times 10^7 s$

now we know that angular speed of the Earth about Sun is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{3.15 \times 10^7}$

now speed of center of Earth is given as

$v_{cm} = r\omega$

$r = 1.5 \times 10^{11} m$

$v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})$

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

now transnational kinetic energy of center of Earth is given as

$K_{trans} = \frac{1}{2}mv^2$

$K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2$

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

Angular speed of Earth about its own axis is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

Now moment of inertia of Earth about its own axis

$I = \frac{2}{5}mR^2$

$I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2$

$I = 9.83 \times 10^{37} kg m^2$

now rotational energy is given as

$KE_{rot} = \frac{1}{2}I\omega^2$

$KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2$

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

Now total kinetic energy is given as

$KE_{tot} = KE_{trans} + KE_{rot}$

$KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}$

$KE_{tot} = 2.68 \times 10^{33} J$

6 0

3 years ago