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3 years ago

14

Accorrding to Kepler's third law, a planet whose distance from the Sun is 2 A.U. would have an orbital period of how many Earth-

years

1 answer:

Lilit [14]3 years ago

4 0

Answer:

hi

Explanation:

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A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s

asambeis [7]

Answer:

$I=0.0987kg.m^2$

Explanation:

From the question we are told that:

Mass $M=1.80kg$

Deviation $d=0.250$

Time $t=0.940s$

Generally the equation for moment of inertia is mathematically given by

$I=\frac{T}{2\pi}^2(mgd)$

$I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)$

$I=0.0987kgm^2$

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3 years ago

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

11111nata11111 [884]

Answer:

$1.62\times 10^{-8}\ \text{s}$

Explanation:

$\epsilon_0$ = Vacuum permittivity = $8.854\times 10^{-12}\ \text{F/m}$

$A$ = Area = $10\times 2\times 10^{-4}\ \text{m}^2$

$d$ = Distance between plates = 1 mm

$V_c$ = Changed voltage = 60 V

$V$ = Initial voltage = 100 V

$R$ = Resistance = $1000\ \Omega$

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}$

We have the relation

$V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}$

The time taken for the potential difference to reach the required level is $1.62\times 10^{-8}\ \text{s}$ .

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2 years ago

What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,

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Both are constants used in the definition of Forces (gravitational and electric,respectively)

Since those constants are proportional to the magnitude of the forces:

Having a small gravitational constant explains why there is no apparent force of attraction with objects of considerable low mass (they would need to have great value of mass for the equation to give an apreciable force)

Electrical interactions are usually strong, and thus require an appropiate constant to depict the phenomenon. We deal in this case with charges really small, but the forces are in different order of magnitude.

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