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AURORKA [14]
3 years ago
9

Which of The following has the most kinetic energy a 25-kg fish tank sitting on a table, A 50-kg Fish swimming in a fish tank, A

7,500-kg car parked on a steep hill, or a 50-kg boulder suspended from a cliff .
Physics
1 answer:
suter [353]3 years ago
3 0

Well, kinetic energy is energy of motion, and three of these
four objects are just sitting there and not moving at all.

The tank, the car, and the boulder have no kinetic energy at all,
since they're not moving.

The fish who is swimming is the only one with any kinetic energy,
so we don't even have to calculate how much.  The fish is your man.
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Explanation:

It is given that,

Wavelength, \lambda_1=500\ nm=5\times 10^{-7}\ m

Wavelength, \lambda_2=500.10\ nm=5.001\times 10^{-7}\ m

We need to find the frequencies from corresponding wavelengths. The frequency of the light is given by :

f=\dfrac{c}{\lambda}  

c is the speed of light

Frequency 1,

f_1=\dfrac{c}{\lambda_1}  

f_1=\dfrac{3\times 10^8\ m/s}{5\times 10^{-7}\ m}  

f_1=6\times 10^{14}\ Hz

Frequency 2,

f_2=\dfrac{c}{\lambda_2}  

f_1=\dfrac{3\times 10^8\ m/s}{5.001\times 10^{-7}\ m}  

f_1=5.99\times 10^{14}\ Hz

Hence, this is the required solution.

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red dwarf stars (stars between 0.08 and 0.5 solar masses) evolve very differently than other stars as they age because ____.
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Red dwarf stars evolve very differently than other stars as they age because <u>their </u><u>interiors </u><u>are well mixed, through strong convection.</u>

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2 years ago
A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial
V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF=2.1\times 10^{-12} F

1pF=10^{-12} F

d=5.0 mm=5\times 10^{-3} m

1mm=10^{-3} m

a.Maximum flux

\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}

\epsilon_0=8.85\times 10^{-12}

\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm

b.Maximum displacement current,I=\frac{V}{R}=\frac{30}{130}=0.23 A

c.We have to find electric flux at t=0.5 ns

t=0.5ns=0.5\times 10^{-9} s

1ns=10^{-9}s

q=CV(1-e^{-\frac{t}{RC}})

q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})

q=52.9\times 10^{-12} C

\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm

d.Displacement current at t=0.5ns

I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}

I=0.037 A

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3 years ago
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