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4 years ago

SOMEONE PLEASE HELP!

Physics

1 answer:

schepotkina [342]4 years ago

4 0

Answer:

I THINK option A is correct

Explanation:

because the electrostatic force is directly proportional to product of charges

now if the force between two electron is F charge on electron is same so their product is positive now the charge on proton is +ve and on electron is -ve so their product is -F

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Which is an example of a mixture?

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Air is the example of mixture

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3 years ago

Which of the following is NOT an insulator?

sergejj [24]

I don’t know how air would be an insulator so I’m guessing that one isn’t

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3 years ago

Read 2 more answers

A parallel RLC resonant circuit has a resistance of 200 Ω. If it is known that the bandwidth is 80 rad/s and the lower half-powe

d1i1m1o1n [39]

Answer:

L= 3.6mH

C =9.9 microfarad

Explanation:

Resonant frequency fr

fr = fl + 1/2 BW

fr = 800+ 1/2×80

=800+40

=840 rad/s

Bandwidth (BW)

BW = fr/Q

Q = quality factor

Q= fr/ BW

Q = 840/80

Q= 10.5

Quality factor = R/Xl

Xl = inductive reactance

Xl = R/Q

Xl = 200/10.5

Xl = 19.05 ohms

Xl =2πfL

L= Inductance

L = Xl /2πf

L =19.05/5278.56

L= 3.6mH

Capacitor C

1= fr^2 × 4π^2LC

C = 1/fr^24π^2L

C = 1/100307.5

C= 9.9microfarad

5 0

3 years ago

PLZ HELP!! List properties of ions.

Andreas93 [3]

Ionic compounds typically have high melting and boiling points, and are hard and brittle. when melted or dissolved they become highly conductive, because the ions are mobilized.

3 0

3 years ago

A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of

SCORPION-xisa [38]

Answer:

a).β=0.53 $x10^{-3}$ T

a).β=0.40 $x10^{-4}$ T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

$\beta =\frac{u_{o}*I*N}{2\pi*r}$

a).

$N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A} \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T$

b).

The distance is the radius add the cross section so:

$r_{1}=15cm+5cm\\r_{1}=20cm$

$r_{1} =20cm*\frac{1m}{100cm}=0.20m$

$\beta =\frac{u_{o}*I*N}{2\pi*r1}$

$\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T$

3 0

3 years ago