The earth is 4.54 billion years old.
Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
Answer:D
Explanation:according to the law of conservation of energy/momentum, when two bodies collides, their total momentum and energy before and after collision are equal. Given that the two bodies move with the same velocities after collision, means that the law has not been violated since momentum = mass x velocity (where mass is constant)
Answer:
W = 30 J
Explanation:
given,
Work done = 10 J
Stretch of spring, x = 0.1 m
We know,
dW = F .dx
we know, F = k x
![W = k[\dfrac{x^2}{2}]_0^{0.1}](https://tex.z-dn.net/?f=W%20%3D%20k%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_0%5E%7B0.1%7D)
k = 2000
now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.
![W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx](https://tex.z-dn.net/?f=W%20%3D%202000%20%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_%7B0.1%7D%5E%7B0.2%7D%20dx)
W = 1000 x 0.03
W = 30 J
Hence, work done is equal to 30 J.
Answer:
Electric field by charged disk is given as
E = (Charge Density/2u0)*[1 - (z/sqrt(z^2 - R^2))]
R = 9.54cm = 0.0954m, z = 1.01m, Charge density = 4.07 x 10^-6C/m2, e0 = 8.85 x 10^-12F/m.
Substituting all the values in to equation,
E = (2.299 x 10^5) x (8.931 x 10^-3)
E = 2.053 x 10^3N/C
Explanation:
