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Gwar [14]
3 years ago
12

How would I workout question 4?? Please help

Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

550 kg

Explanation:

mass = E / gh

= 33000/60

=550

plzzz......

mark it as a brilliant answer

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A parallel plate capacitor is charged to a potential difference of 100 V and disconnected from the source. A slab of dielectric
choli [55]

Answer:

Check Explanation

Explanation:

a) The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U₀ = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

ϵ = kϵ₀

where k = dielectric constant

ϵ₀ = permissivity of free space/air/vacuum

A = Cross sectional Area of the capacitor

d = separation between the capacitor

If air/vacuum/free space are the dielectric constants,

So, k = 1 and ϵ = ϵ₀

U₀ = CV²/2

Substituting for C

U₀ = ϵ₀AV²/2d

The slab of dielectric is now inserted in the space between the plates of the capacitor,

The dielectric material has a dielectric constant of k

ϵ = kϵ₀

U = (kϵ₀AV²)/2d

Compared to U₀

U = (kϵ₀AV²)/2d

U₀ = (ϵ₀AV²)/2d

(U/U₀) = k (Proved)

b) The dielectric constant of a dielectric material is an expression that shows how much the material concentrates the electric flux between the plates of the capacitor. As it is a ratio, it compares this ability with the ability of air/vacuum/free space to concentrate the required electric flux.

So, any material with a dielectric constant greater than 1 has the ability to enable the capacitor to store more charges, thereby leading to more energy stored in that capacitor.

c) As shown in (a), the capacitance, C is directly proportional to the absolute permissivity of the dielectric material between the plates of the capacitor. So, as the dielectric constant of the dielectric material increases, the capacitance of the the capacitor increases as long as the O the parameters stay constant.

8 0
3 years ago
a stone is vertically thrown upward with the velocity of 72km/hr find the maximum height reached the height​
erik [133]

Answer:

y = 20.38 [m]

Explanation:

In order to solve these problems, we must use the following kinematics equation.

v_{f} ^{2} =v_{i} ^{2}-(2*g*y)

where:

Vf = final velocity = 0

Vi = initial velocity = 72 [km/h]

g = gravity acceleration = 9.81 [m/s^2]

y = vertical elevation [m]

We need to convert [km/h] to [m/s]

72[\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km} ] = 20 [m/s]

Note: the negative sign of the equation means that the acceleration acts in the opposite direction to the movement of the body. And the final speed is zero, because when the body reaches the maximum height, the Stone does not move its speed has been reduced to its entirety.

0 = (20)^2 - (2*9.81*y)

20^2 = 2*9.81*y

y = 20.38 [m]

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Where a major earthquake has happened in the past, but not in a while, is called a seismic _____.
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It is called a seismic gap.

A.
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