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kykrilka [37]
3 years ago
6

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

<u>Physical</u><u> </u><u>quantity</u><u>:</u><u>-units</u>

  • 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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Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

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This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

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