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Zanzabum
3 years ago
15

Current is the movement of positive charges called electrons

Physics
2 answers:
enot [183]3 years ago
8 0

Answer:yes

Explanation:current is the movement of electrons along a conductor

GaryK [48]3 years ago
6 0

Answer:what

Yeah

Explanation:

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When minerals are exposed to high pressure and temperatures, the individual minerals begin to realign themselves
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Describe a situation in which an electron will be affected by an external electric field but will not be affected by an external
irina [24]

Answer:

Explanation:

Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field

a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.

b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will  exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.

Situations when electron is affected by an external magnetic field but not by an external electric field

There is no such situation in which electric field will not affect an electron . It will always affect an electron .

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What is the name of Newton's first law of motion? A. law of inertia B. law of acceleration C. law of objects at rest D. law of o
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The answer is a.......
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3 years ago
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Which properties do metalloids share with metals?
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A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
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