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Trava [24]
2 years ago
12

What is the advantage of the SI unit over CGS units?

Physics
1 answer:
MatroZZZ [7]2 years ago
8 0

Answer:

The advantage of the SI unit over CGS unit are:

  • S.I has broader base. It has seven base units and two supplementary units.
  • S.I is metric.
  • S.I is coherent.
  • S.I is rational, i.e. it gives one unit for one physical quantity .e g. for energy of any type i.e, mechanical or heat or electrical.There is only one unit Joule(J) but in M.K.S system unit for mechanical energy is joule.

<em><u>HOPE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>WILL</u></em><em><u> </u></em><em><u>HELP</u></em><em><u> </u></em><em><u>YOU</u></em>

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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-c
AleksAgata [21]

Answer:

Explanation:

Let L be the length of the wire.

velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L  m /s

mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m

m = 7.25 x 10⁻⁷ kg / m

Tension in the wire = Mg  , M is mass hanged from lower end.

= .4 x 9.8

= 3.92 N

expression for velocity of wave in the wire

v = \sqrt{\frac{T}{m} }    , T is tension in the wire , m is mass per unit length of wire .

40.48 L = \sqrt{\frac{3.92}{7.25\times10^{-7}} }

1638.63 L² = 3.92 / (7.25 x 10⁻⁷)

L² = 3.92 x 10⁷ / (7.25 x 1638.63 )

L² = 3299.64

L = 57.44 m /s

5 0
2 years ago
A car drives around a racetrack for 30 seconds. what do you need to know to calculate the average velocity of the car?
boyakko [2]
The time is given, and you want to find the average velocity. To do this, you need to know the distance covered by the driver around the racetrack in that 30 seconds. You divide this by the time, then you will obtain the average velocity in units of, say meters per second.
8 0
2 years ago
Read 2 more answers
A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the
zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂=  L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0
2 years ago
Read 2 more answers
Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as
Sonja [21]

Explanation:

The waveform expression is given by :

y(x,t)=y_m\ sin(0.333x+5.36+585t)...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

y(x,t)=y_m\ sin(kx+\phi+\omega t)..........(2)

Where

k=\dfrac{2\pi}{\lambda}

\omega=2\pi f

On comparing equation (1) and (2) we get :

0.333=\dfrac{2\pi}{\lambda}

\lambda=18.86\ m

585=2\pi f

f = 93.10 Hz

Time period, T=\dfrac{1}{f}

T=\dfrac{1}{0.010}

T = 0.010 s

Phase constant, \phi=5.36\ radian

Hence, this is the required solution.

8 0
2 years ago
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