Answer:
ΔVab = Ed
ΔVab = Va-Vb = Va-V0 = Va
E = Va/ d
= 413V / 0.0795 m
= 5194.97 V/M
Explanation:
the potential difference between two uniform plates is calculated by the formula of electric field.
Answer:
The work done by a particle from x = 0 to x = 2 m is 20 J.
Explanation:
A force on a particle depends on position constrained to move along the x-axis, is given by,
We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.
We know that the work done by a particle is given by the formula as follows :
So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Answer:
The duration is ![T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m](https://tex.z-dn.net/?f=T%20%20%3D72%20%5C%20%20years%20%2Ftex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EFrom%20the%20question%20we%20are%20told%20that%20%3C%2Fp%3E%3Cp%3E%20%20%20%20The%20%20distance%20is%20%20%5Btex%5DD%20%20%3D%20%2035%20%5C%20light-years%20%3D%2035%20%2A%20%209.46%20%2A10%5E%7B15%7D%20%3D%203.311%20%2A10%5E%7B17%7D%20%5C%20%20m%20)
Generally the time it would take for the message to get the the other civilization is mathematically represented as
Here c is the speed of light with the value 
=> 
=> 
converting to years
Now the total time taken is mathematically represented as
=> 
=> [tex]T =72 \ years /tex]
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
