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True or False:

mart [117]

This is a true statement

3 0

3 years ago

Read 2 more answers

Adding more pulleys to a system of pulleys _____ needed to lift an object.

-Dominant- [34]

I think it’s D-decreases the amount of work.

3 0

3 years ago

g Where is the dielectric constant and A is the area of the capacitor plates. The charge q and voltage Vc across the capacitor p

aev [14]

Answer:

Capacitor and the dielectrics:

"The capacitor is a charge,q storing device as it is subjected to the potential difference between the different plates and thus results in the difference of charge decomposition between the two plates."

Formula:C= ∈°(A/d),

Unit: Farad,F.

Purpose of the Dielectrics:

In order to store or hold the optimum number of charge particles,q between the plates we have a number of insulators which are used as the medium between the two plates.While, it separates both the plates from one another.

Explanation:

Dielectric Constants in the capacitors:

In order to hold the number of charges in between the plates of the capacitor a specific medium or substrate is placed(mostly the insulators). As it covers the area between the two capacitors, while it can decrease and increase by the action of the movable plate of the capacitor.

Now, it increase the capcitative power of the capacitor by decreasing the potential difference between the two plates, along with which the electromagnetic field is also greatly reduced.As the interaction between the two plates is greatly reduced, as they are covered and well protected by the sheet or membrane of insulting medium around them.

6 0

3 years ago

A planet has a gravitational acceleration on its surface of 2.2 times Earth's gravitational acceleration on its surface. The pla

lesantik [10]

Answer:

The mass of the planet is 55 times the mass of earth.

Explanation:

From the inverse-square gravitation law,

F = (GMm/r²)

If the weight of a body (the force with which the earth attracts a body to its centre) is to be calculated,

F = mg

m = mass of the body,

g = acceleration due to gravity

mg = (GMm/r²)

G = Gravitational constant

M = mass of the earth

m = mass of body

r = distance between the body and the centre of the earth = radius of the earth

The acceleration due to gravity is given by

g = (GM/r²)

Making the mass of the earth, the subject of formula

M = (gr²/G) (eqn 1)

So, the planet described,

Let the acceleration due to gravity on the planet be g₁

Mass of the planet be M₁

Radius of the planet be r₁

g₁ = 2.2g

r₁ = 5r

M₁ = ?

Note that the gravitational constant is the same for both planets.

So, we can write a similar expression for the planet's acceleration due to gravity

g₁ = (GM₁/r₁²)

Substituting all the parameters known in terms of their corresponding earth values

2.2g = [GM₁/(5r)²]

2.2g = [GM₁/25r²]

M₁ = (55gr²/G)

Recall the expression for the mass of the earth

M = (gr²/G)

M₁ = 55 M

The mass of the planet, in terms of Earth masses = 55M

The mass of the planet is 55 times the planet of earth.

Hope this Helps!!!

5 0

3 years ago

A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axi

KiRa [710]

Answer:

The induced voltage is $\epsilon = 4.53 \ V$

Explanation:

From the question we are told that

The number of turns is $N = 1300 \ turns$

The diameter is $d = 2.2 \ cm =0.022 \ m$

The initial magnetic field is $B_i = 0.11 \ T$

The final magnetic field is $B_f = 0 \ T$

The time taken is $t = 12 \ ms = 12*10^{-3} \ s$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{0.022}{2}$

$r = 0.011 \ m$

Generally the induced emf is mathematically represented as

$\epsilon = - N * \frac{d\phi}{dt}$

Where $d\phi$ is the change in magnetic flux of the wire which is mathematically represented as

$d \phi = dB* A * cos \theta$

=> $d \phi = (B_f - B_i )* A * cos \theta$

Here $\theta = 0$

since the axis of the coil is parallel to the field

Where A is the cross-sectional area of the coil which is mathematically represented as

$A = \pi * r^2$

$A = 3.142 * 0.011^2$

$A = 3.80*10^{-4} \ m^2$

So the induced emf

$\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}$ Here we substituted the values of $d \phi$

$\epsilon = 4.53 \ V$

6 0

3 years ago