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3 years ago

A 1.5 m wire carries a 9A current when a potential difference of 84 v is applied. what is the resistance of the wire

2 answers:

8 0

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Ann [662]3 years ago

3 0

Answer : The resistance of the wire is, 9.33 ohms

Solution : Given,

Electric current = 9 A

Potential difference = 84 V

Using ohm's law,

$V=I\times R$

or,

$R=\frac{V}{I}$

where,

R = resistance of wire

V = potential difference

I = electric current

Now put all the given values in the above formula, we get the resistance of the wire.

$R=\frac{84V}{9A}=9.33\Omega$

Therefore, the resistance of the wire is, 9.33 ohms

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The speed of light is 3.0x10^5 km/sec and it takes light 1.3 seconds for light to travel from the moon to earth froim this infor

Vika [28.1K]

The motion of the light is a uniform motion with constant speed $v=3 \cdot 10^5 km/s$ , therefore we can use the basic relationship between speed, space and time:
$v= \frac{S}{t}$ (1)
where S is the distance covered and t is the time taken. The light takes t=1.3 s to travel from the moon to Earth, therefore by rearranging eq.(1) we can find the distance between the Moon and the Earth:
$S=vt=(3 \cdot 10^5 km/s)(1.3 s)=390000 km=3.9 \cdot 10^5 km$

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3 years ago

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0

3 years ago

A hydrogen atom has 1 electron. In which level is it?

kolezko [41]

If it is in its ground state, it will be in the level 1s

3 0

3 years ago

If the particle has mass m, how fast must it be moving away from the Sun's center of mass to escape the gravitational influence

IgorLugansk [536]

Answer:

Explanation:

M = 1.989 x 10^30 kg

R = 6.96 x 10^8 m

G = 6.67 x 10^-11 Nm²/kg²

Let the velocity is v.

$v=\sqrt{\frac{2GM}{R}}$

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}}{6.96\times 10^{8}}}$

v = 6.17 x 10^5 m/s

3 0

3 years ago

if you throw a ball straight up into the air at a velocity of 15 m/s you want to know how high above your hand the ball will be

xxMikexx [17]

Answer:

Assuming that gravity has not caused the ball to start falling then after 2.5 seconds the ball will be 37.5 meters in the air

Explanation:

The ball travels 15 meters every second so for 2 seconds we would multiply 15*2=30. We have 2.5 seconds so to calculate the time traveled in half a second we would divide 15/2=7.5

Then we add 30+7.5= 37.5 meters

4 0

2 years ago