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Basile [38]
3 years ago
5

An object with charge q=−6.00×10−9C is placed in a region of uniform electric field and is released from rest at point A. After

the charge has moved to point B, 0.500 m to the right, it has kinetic energy 3.00×10−7J.
(a) If the electric potential at point A is +30.0 V, what is the electric potential at point B?
(b) What are the magnitude and direction of the electric field?
Physics
1 answer:
shtirl [24]3 years ago
8 0

Answer:

its a

Explanation:

used google

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Nikolay [14]

15 min

Explanation:

take 0.25 and put it in for 1.00 and you will see its 0.25 but when you add it all 4 times it is 1.00 so then you would take that and do it to the hour ... how many times does four go into 60

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What does the voltage-current graph above show about the relationship between voltage and current, and how do these properties a
zubka84 [21]

The answer is as voltage increases current increases and therefore resistance would remain constant

4 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
Monochromatic light with a wavelength of 600 nanometers (one nanometer is 10-9 meters) is incident upon a double slit arrangemen
AnnyKZ [126]
I think it should be 6,000
7 0
2 years ago
At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa
Elan Coil [88]

Answer:

2. west

Explanation:

Given an electron is moving vertically upward from ground.

Now Fleming right hand rule state that: make L shape with thumb and index finger then point middle finger perpendicular to index and thumb.Then index finger points in the direction of moving charge , middle finger points in the direction of the magnetic field and thumb points in the direction of the magnetic force.

According to Fleming right hand rule the direction of the magnetic that acts on the electron is west.

5 0
3 years ago
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