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lozanna [386]
3 years ago
9

While solving a problem, Tran calculates an answer that has coulomb-volts as the units.

Physics
1 answer:
Bess [88]3 years ago
4 0

Apparently, the question is looking for A. electric potential energy;

but I don't think that's quite right. Electric potential difference is expressed in Joules / Coulomb which is the work to move a charge between 2 points

Example: If the electric field between, say, between 2 capacitor plates is

E = 100 Newtons / Coulomb then the work done in moving a unit of charge from the negative plate to the positive plate separted by 1 cm is

V = E * d = 100 Newtons / Coulomb * .01 meters = 1 Newton-meter / Coulomb

= 1 Joule / Coulomb    which is the electric potential or potential difference

(The definition of electric potential between points is "the work moving a unit positive test charge from one point to the other")

Now in our above example where V = 1 Joule / Coulomb

if we move 10 Coulombs from the negative plate to the positive plate

W = V Q  = 1 Joule / Coulomb * 10 Coulombs = 10 Joules

where work done has the correct units of Joules.

Your textbook should help clarify this.

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At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?
kompoz [17]

Answer:

\lambda=4.81\times 10^{-8}\ m

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

\lambda T=2.89\times 10^{-3}

Here,

\lambda = wavelength

\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m

So, the wavelength of the star is 4.81\times 10^{-8}\ m.

7 0
2 years ago
Describing a Physical Change
babymother [125]

Answer:

physical change is the temporary change or riversible change here the physical properties r only changed

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6 0
2 years ago
An object that travels around another object in space is called a(n)
HACTEHA [7]
Satellite. think of the moon.

Hope this helps!
Vote me Brainliest!
3 0
3 years ago
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magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
What formula is needed to calculate the acceleration of object in motion?
Butoxors [25]

Answer:

Acceleration = Force/Mass

Explanation:

6 0
3 years ago
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