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Nataliya [291]
3 years ago
12

A capacitor is connected to an AC generator. As the generator's frequency is increased, what happens to the current in the capac

itor
Physics
1 answer:
Licemer1 [7]3 years ago
8 0

Answer:

The current increases.

Explanation:

A capacitor can be defined as an electronic component used in electrical circuits to store charge temporarily.

A capacitor is connected to an AC generator. As the generator's frequency is increased, the current in the capacitor increases as well.

This ultimately implies that, when a capacitor is connected to an AC generator, the frequency of the circuit is directly proportional to the amount of current flowing through it.

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2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would
professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

Circumference=2\pi R

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

v=\frac{Distance}{Time}

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

5 0
3 years ago
The inductance of a closely packed coil of 560 turns is 8.9 mH. Calculate the magnetic flux through the coil when the current is
Y_Kistochka [10]

Answer:

0.11\times 10^{-6}weber

Explanation:

We have given number of turns N = 560

Inductance L = 8.9 mH

Current through the coil = 7 mA

Inductance of the coil is given as L=\frac{N\Phi }{I}

Where N  is number of turns I is current and \Phi is flux

So \Phi =\frac{LI}{N}=\frac{8.9\times 10^{-3}\times 7\times 10^{-3}}{560}=0.11\times 10^{-6}weber

6 0
3 years ago
Read 2 more answers
All objects emit ______ radiation?<br> A-electromagnetic<br> B-kinetic<br> D-solar
natulia [17]
A, electromagnetic radiation
6 0
3 years ago
Iodine molecules separates into atoms after absorbing light of 450 nm. If one
OlgaM077 [116]

Answer:E = hc/? = 4.41 x 10-19 J

Energy absorbed by each atom :

E (atom) = 2.205 x 10-19 J

Now Bond Energy of each molecule (B) = 3.98 x J

So, for each atom 1.99 x 10-19 J

So now

KE of each atom = E(atom) - B (atom)

= 2.15 x 10-19 J

5 0
3 years ago
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