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2 years ago

A 210 Ohm resistor uses 9.28 W of

Physics

1 answer:

IgorLugansk [536]2 years ago

4 0

Resistance=R=210Ohm
Power=9.28W=P

Current=I

$\boxed{\sf P=I^2R}$

$\\ \sf\longmapsto I^2=\dfrac{P}{R}$

$\\ \sf\longmapsto I^2=\dfrac{9.28}{210}$

$\\ \sf\longmapsto I^2\approx0.04$

$\\ \sf\longmapsto I\approx\sqrt{0.04}$

$\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}$

$\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}$

$\\ \sf\longmapsto I\approx\dfrac{2}{10}$

$\\ \sf\longmapsto I\approx0.2A$

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Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea

Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ( $m_{E}$ , $m_{M}$ ) are $5.972\times 10^{24}\,kg$ and $7.349\times 10^{22}\,kg$ , respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

$\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}$

If $x_{E} = 0\,km$ and $x_{M} = 384,403\,km$ , then:

$\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}$

$\bar x = 4.673\,km$

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

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3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

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3 years ago

A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r

bazaltina [42]

Answer:

$V=16.65 m^3$

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

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So the net force uproad on the balloon is

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Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

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$F_b=F_A-F_H$

$F_b=m_a*g-m_h*g$

$m=P/V$

$18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g$

$V=\frac{18.5kg}{(1.29-0.179)kg/m^3}$

$V=16.65 m^3$

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3 years ago

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Ivahew [28]

Answer:

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Explanation:

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3 years ago

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If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has be

ale4655 [162]

Answer:

500J

Explanation:

The arrow will have an energy of 500J after it has been released from its state of rest.

This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".

The energy at rest which is the potential energy is 500J
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This way, no energy is lost and we can account for the energy transformations occurring.

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2 years ago