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Ulleksa [173]
4 years ago
6

If the potential across two parallel plates, separated by 3 cm, is 12 volts, what is the electric field strength in volts per me

ter? E = _____ volts/m
Physics
2 answers:
BARSIC [14]4 years ago
8 0

Answer:

E = 400 V/m

Explanation:

It is given that, for a parallel plates :

Potential difference, V = 12 volts

Separation between the plates, d = 3 cm = 0.03 m

The relation between the electric field and the electric potential is given as :

E=\dfrac{V}{d}

E=\dfrac{12\ V}{0.03\ m}

E = 400 volts/m

So, the strength of the electric field is 400 V/m.                

Mrrafil [7]4 years ago
6 0
There are 100cm in 1m, divide 100 by 3 and you get 33, multiply that by 12 and you get  396 volts/m.

Hope this is correct and helps.
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What element x is most likely to react to form the compound xf5?
djyliett [7]
<span>antimony. It has +3,+5,-3 so yeah. the others carbon+2,+4,-4, chlorine +1,+5,+7,-1 but -1 is the most often so it isn't Cl, calcium +2.</span>
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4 years ago
Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on
mrs_skeptik [129]

Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

dU = -2.59x10^4 - 6.46^4

dU = -9.05x10^4 kJ

8 0
3 years ago
Calculate the number of moles of water molecules in 12 dm' of water<br>vapour at STP.<br><br>​
Vinvika [58]

Answer:

22.4 \:  {dm}^{3}  \: are \: occupied \: by \: 1 \: mole \\ 12 \:  {dm}^{3}  \: will \: be \: occupied \: by \: (  \frac{12}{22.4} ) \: moles \\  = 0.536 \: moles

4 0
3 years ago
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0
3 years ago
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
3 years ago
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