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2 years ago

Write 20 physical Quantities withS.I unit and formula

Physics

2 answers:

Svetach [21]2 years ago

7 0

Answer:

HERE IS YOUR ANSWER

Physical quantity => Unit

1)Length=>metre

2)Time=>second

3)Temperature=>Kelvin

4)mass=>kilogram

5)electric current=>ampere

6)luminous intensity=>candela

7)amount of substance=>mole

8)velocity=>m/s

9)acceleration=>m/s²

10)momentum=>kgm/s

11)density=>kg/m³

12)volume=>m³

13)force=>Newton(N)

14)Energy=>Joule(J)

15)Power=>Watt(W)

16)Pressure=>Pascal(Pa) or N/m²

17)Resistance=>ohm

18)Electrical potential=>volt(V)

19)plane angle=>radian

20)solid angle=>steradian

Explanation:

Ainat [17]2 years ago

6 0

Physical quantity:-units

Force, Weight:-Newton

Frequency:-Hertz.

Electric charge:-Coulomb

Electric potential (Voltage):-Volt.

Inductance:-Henry

.Capacitance:-Farad

Resistance, Impedance, Reactance:-Ohm

Electricalconductance:-Siemens.

Magneticflux:-Weber

Magnetic flux density:-Tesla.

Energy, Work, Hea:- Joule

Power, Radiant flux:-Watt

Angle:- Radian

Radioactivity :-Becquerel

Luminous flux:-leman

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An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10 µC that does not touch the

aliina [53]

Answer:

Explanation:

we have to make charge inside the conductor zero because we know that electric field inside the conductor should be zero

so, the outer surface of the conductor should contain + 10 uC of charge and the inner surface contains -10 uC

8 0

2 years ago

The electric potential at the dot in the figure is 3160 V. What is charge q?

PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

5 0

2 years ago

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50

Setler79 [48]

Answer:

very hard others will answer it

Explanation:

hard

6 0

2 years ago

A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)

bezimeni [28]

Answer:

205 V

V $_{R}$ = 2.05 V

Explanation:

L = Inductance in Henries, (H) = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

$V_{L} = - IwLsin(wt)$

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

= 11.0 V / 500 rad/s (0.500 H)

= 11.0 / 250

I = 0.044 A

Now

V $_{R}$ = IR

= (0.044 A) (93 Ω)

V $_{R}$ = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V $_{R}$ = V $_{R}$ cos (wt)

Putting V $_{R}$ = 4.092 V and w = 500 rad/s

V $_{R}$ = V $_{R}$ cos (wt)

= (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V $_{R}$ = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

= (4.092 V) (cos (500 rads/s)(0.00209))

= (4.092 V) (cos(1.045))

= (4.092 V)(0.501902)

= 2.053783

V $_{R}$ = 2.05 V

8 0

2 years ago

I need help fast please help

Irina18 [472]

A and C Im pretty sure :)

5 0

2 years ago

Write 20 physical Quantities withS.I unit and formula​

Write 20 physical Quantities withS.I unit and formula