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3 years ago

Write 20 physical Quantities withS.I unit and formula

Physics

2 answers:

Svetach [21]3 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]3 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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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an

LUCKY_DIMON [66]

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

$\Phi_B=BA=\mu_oNIA$

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

$A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\$

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

$M=\mu_o N_1 N_2 \frac{A_2}{l}$

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

$M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H$

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

$\epsilon_1=M\frac{dI_2}{dt}$

by replacing you obtain:

$\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV$

the emf is 20mV

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3 years ago

Write 20 physical Quantities withS.I unit and formula​

Write 20 physical Quantities withS.I unit and formula