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AlladinOne [14]
3 years ago
7

Si unit of small g and G​

Physics
1 answer:
Mila [183]3 years ago
5 0

Answer:

G

Explanation:

because the Si unite of g is G mean gram

You might be interested in
A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor
OLEGan [10]

Answer:

(e)6.835\times 10^{-7}F

Explanation:

At resonance we know that X_l=X_C

That is \omega L=\frac{1}{\omega C}

\omega ^2=\frac{1}{LC}

\omega =\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So 4511=\frac{1}{2\pi \sqrt{LC}}

LC=\frac{1}{4\pi ^2\times 4511^2}

LC=1.244\times 10^{-9}

C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F

So option e is the correct answer

3 0
3 years ago
At what distance along the z-axis is the electric field strength a maximum?
Lesechka [4]
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is 

<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>

<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>

<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>

<span>so the equation becomes </span>

<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>

<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>

<span>Integrating around the ring you get </span>

<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>

<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>

<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>

<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>

<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>

<span>to find the maxima set this = 0, giving </span>

<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>

<span>mult both side by (x^2 + R^2)^2.5 to get </span>

<span>(x^2 + R^2) - 3*x^2 = 0 </span>

<span>-2*x^2 + R^2 = 0 </span>

<span>-2*x^2 = -R^2 </span>

<span>x = (+/-)R/sqrt(2) </span>
4 0
3 years ago
A 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. Calculate the
kumpel [21]

The car’s momentum after 4.21s is 24617.4 kgm/s

<h3>Newton's Second Law of Motion.</h3>

Newton's second law state that, the rate of change of momentum, is directly proportional to the applied force.

Given that a 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. To calculate the car’s momentum after 4.21 s, Let us first list all the parameters involved.

  • Velocity u = 10.2 m/s
  • Acceleration a =  2.45 m/s²
  • Mass m = 1200Kg
  • Time t = 4.21 s

From Newton's second law,

F = (mv - mu) / t

ma = (mv - mu) / t

Substitute all the parameters into the formula above.

1200 × 2.45 = ( mv - 1200 × 10.2 ) / 4.21

2940 = ( mv - 12240 ) / 4.21

Cross multiply

12377.4 = mv - 12240

Make mv the subject of the formula

mv = 12377.4 + 12240

mv = 24617.4 kgm/s

Therefore, the car’s momentum after 4.21s is 24617.4 kgm/s

Learn more about Momentum here: brainly.com/question/25121535

#SPJ1

3 0
1 year ago
Bill was really bored at work. During his break, he decided to check his carotid artery. He placed both fingers on his neck and
Sergeeva-Olga [200]

Answer:

72 beats per minute

Explanation:

Heart beat causes the flow of blood round the body. This heart beat can be felt as pulse in the wrist or neck carotid artery. The heart rate which is measured in beats per minute (BPM) is used to determine the number of heart beats per minute.

You can calculate your BPM using the carotid artery found in the neck close to the windpipe.

Given that for 20 seconds, Bill had a total of 24 beats.

60 seconds = 1 minute.

Hence,  Bill's BPM = (24 beats per 20 seconds) * (60 seconds per minute) = 72 beats per minute

5 0
2 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
vovikov84 [41]

Answer:

The 50-W bulb glows more brightly than the 100-W bulb

Explanation:

The bulb has a rating of 100 W under 110 V . So it will glow with full brightness when it is fed 110 V . When bulbs are in parallel combination , each bulb receives 110 V . So they glow with full brightness .

When they are in series combination , 110 supply voltage gets distributed between the , thus  , reducing the voltage appearing on each of them less than 110 V . So their brightness is reduced.

resistance = V

Bulb having high wattage rating has low resistance resulting in higher current . In the second case , both have same current as they are in series combination . So more heat will be generated in bulb having more resistance . Since 50 W bulb has higher resistance , it will glow brighter than 100 W bulb.

4 0
3 years ago
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