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Tpy6a [65]
3 years ago
10

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unch

anged?
Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

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1. thomas jefferson proposed using the length (l) of a simple pendulum whose period (t) was exactly 2 seconds as the definition
Alecsey [184]

The length of a 2 sec pendulum is 1 m.

Given that, initial length of the simple pendulum L₁ = 1 m

Initial time period T₁ = 2 sec

We need to find the length of the pendulum whose time period is 2 sec

T₂ = 2 sec

L₂ = ?

We know that the time period of the simple pendulum is given by the formula,

T = 2π√(L/g)

From the above relation, we can write T ∝ √L

T₁ / T₂ = √(L₁/L₂)

Making L₂ from the above relation, we have,

L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m

Thus, the length of a 2 sec pendulum is 1 m.

To know more about time period:

brainly.com/question/17350379

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7 0
1 year ago
Which product of nuclear decay has mass but no charge?
Tom [10]
Neutrons have a zero charge but consist of mass.
3 0
3 years ago
What is the kinetic energy of an object that has a mass of 50.0 kg and a velocity of 18 m/s? A.450 J B.900 J C.8,100 J D.16,000
Svet_ta [14]
KE = (1/2)·(mass)·(speed)²

KE = (1/2)·(50 kg)·(18 m/s)²

KE = (25 kg)·(324 m²/s²)

KE = 8,100 kg-m²/s²

KE = 8,100 Joules
5 0
3 years ago
Read 2 more answers
How are magnetic poles and electrical charges similar? A. Their same forces repel each other, and their opposite forces attract
melomori [17]

Answer:

A. Their same forces repel each other, and their opposite forces attract each other.

Explanation:

As we know by the property of electric charge that similar charges repel each other and opposite charges attracts each other

So here we have similar situation in magnets also

Two similar poles or like poles of magnet i.e. North - North poles or South - South poles always repel each other

And also we can see that opposite poles of magnet each other i.e. North pole and south pole always attract each other.

SO here correct answer is

A. Their same forces repel each other, and their opposite forces attract each other.

8 0
3 years ago
Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci
ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0
3 years ago
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