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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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What produces the distinctive flame color of different substances when they are ignited?
shepuryov [24]

Answer:

Electrons in different metals

Explanation:

When you heat an atom, some of its electrons are "excited* to higher energy levels. When an electron drops from one level to a lower energy level, it emits a quantum of energy.  The different mix of energy differences for each atom produces different colours. Each metal gives a characteristic flame emission spectrum.

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3 years ago
Planet RMM-1 has a mass of 28,500 kg and the star it revolves around has
SOVA2 [1]

Gravitational force = G · (mass₁) · (mass₂) / (distance)

(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)

G = 6.67 x 10⁻¹¹ n-m² / kg²  (the "gravitational constant")

Distance²  = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)

Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)

Distance²  =  (418.209 N-m²) / (39N)

Distance²  =  10.72 m²

<em>Distance = 3.275 meters</em>

An absurd scenario, but that's by golly what the math says with the numbers provided.  I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.

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Social assistance is a synonym for social service.<br> False<br> True
nadezda [96]
The answer is that it is true
6 0
3 years ago
Read 2 more answers
A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

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