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3 years ago

What is time and it si unit

Physics

2 answers:

faust18 [17]3 years ago

5 0

Answer:

Time: The concept of time is self-evident. An hour consists of a certain number of minutes, a day of hours and a year of days. Time is represented through change, such as the circular motion of the moon around Earth. The passing of time is indeed closely connected to the concept of space.

The SI unit for time is second: The second, symbol s, is the SI unit of time. It is defined by taking the fixed numerical value of the cesium frequency ΔνCs, the unperturbed ground-state hyperfine transition frequency of the cesium 133 atom, to be 9 192 631 770 when expressed in the unit Hz, which is equal to s-1.

stich3 [128]3 years ago

3 0

Answer:

the duration between any two events is called time. It's si unit is second.

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Answer: It will remain the same

Explanation:

According to Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them".

Mathematically this law is written as:

$F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}$ is the electrostatic force

$K$ is the Coulomb's constant

$q_{1}=q_{2}=q$ are the electric charges , which in this case have the same positive charge

$d$ is the separation distance between the charges

Rewritting we have:

$F_{E}=K\frac{q^{2}}{d^{2}}$ (2)

Now, if the first charge is doubled:

$q_{1}=2q_$

And the second is reduced to a half:

$q_{2}=\frac{1}{2}q$

We will have the following:

$F_{E}=K\frac{(2q)(\frac{1}{2}q)}{d^{2}}$ (3)

$F_{E}=K\frac{q^{2}}{d^{2}}$ (4)

As we can see equation (4) is equal to equation (2), this means the force of repulsion between both charges will remain the same

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3 years ago

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The brightest star in the night sky in the northern hemisphere is Sirius. Its distance from Earth is estimated to be 8.7 light y

katen-ka-za [31]

Answer: $5.11(10)^{13}miles$

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.

This unit is equivalent to $5.879(10)^{12}miles$ , which mathematically is expressed as:

$1Ly=5.879(10)^{12}miles$

Doing the conversion:

$8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles$ This is the distance from Earth to Sirius in miles.

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Their combined momentum after they meet is 0 .

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3 years ago

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Answer:

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3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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3 years ago

What is time and it si unit​

What is time and it si unit