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mylen [45]
4 years ago
9

The potential difference between the plates of an ideal air-filled parallel-plate capacitor with a plate separation of 6.0 cm is

60 V. What is the strength of the electric field between the plates of this capacitor?
Physics
1 answer:
Yanka [14]4 years ago
8 0

Answer:

1000 N/C

Explanation:

Potential difference, V = 60 V

Distance between the plates, d = 6 cm = 0.06 m

The electric field between the plates is given by

E = V / d

E = 60 / 0.06 = 1000 N/C

Thus, the electric filed between the plates is 1000 N/C.

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Answer:

none

Explanation:

it's to high up to be affected by the gravity

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2 years ago
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A 1400 kg car is moving at 33.8 m/s when a force is applied the opposite direction of the car's motion. The car slows down to 21
zubka84 [21]

Solve for acceleration:

<em>a</em> = (21.4 m/s - 33.8 m/s) / (4.7 s)

<em>a</em> ≈ -2.6 m/s²

Solve for force:

<em>F</em> = (1400 kg) <em>a</em> ≈ -3700 N

The minus sign tells you the force points in the opposite direction of the car's motion. Its magnitude is always positive, so <em>F</em> = 3700 N.

3 0
3 years ago
QUICCCKKKKK!!!!!!!!!!Stimulus discrimination occurs when an organism generalizes one consequence to many stimuli similar to the
timama [110]
False i just took the test and put true as a guess but got it wrong so it is false

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7 0
3 years ago
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If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :
gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

  • c=1

So

  • b-3c=1
  • b-3=1
  • b=1+3
  • b=4

and

  • -a-b=-1
  • -a-4=-1
  • -a=-1+4=3
  • a=-3

So the unit is

  • DV⁴/F³
5 0
2 years ago
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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
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