a) 1.57 m/s
The sock spins once every 2.0 seconds, so its period is
T = 2.0 s
Therefore, the angular velocity of the sock is

The linear speed of the sock is given by

where
is the angular velocity
r = 0.50 m is the radius of the circular path of the sock
Substituting, we find:

B) Faster
In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

Therefore, the new linear speed would be:

And substituting,

So, we see that the linear speed has doubled.
Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; 
Age of the universe; 
We know that, the reciprocal of the Hubble's constant (
) gives an estimate of the age of the universe (
). It is expressed as:

Now,
Hubble's constant; 
We know that;

so
![1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m](https://tex.z-dn.net/?f=1%5C%20Million%5C%20light%5C%20years%20%3D%20%5B9.46%20%2A%2010%5E%7B15%7Dm%5D%20%2A%2010%5E6%20%3D%209.46%20%2A%2010%5E%7B21%7Dm)
Therefore;

Now, we input this Hubble's constant value into our equation;

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Learn more: brainly.com/question/14019680
Answer:
0.31 m
Explanation:
m = mass of the block = 1.5 kg
H = height from which the block is released on ramp = 0.81 m
k = spring constant of the spring = 250 N/m
x = maximum compression of the spring
using conservation of energy
Spring potential energy gained by spring = Potential energy lost by block
(0.5) k x² = mgH
(0.5) (250) x² = (1.5) (9.8) (0.81)
x = 0.31 m
Answer:
V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s average speed
t = 7.2 / 7.25 = .993 sec time to cross patch
a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2 or 1.5 m/s^2
Answer:
= 1220 nm
= 1.22 μm
Explanation:
given data:
wavelength 
distance of screen from slits D = 3 m
1st order bright fringe is 4.84 mm
condition for 1 st bright is
---( 1)
and

= 0.0924 degrees
plug theta value in equation 1 we get


condition for 1 st dark fringe


= 2λ since from eq (1)
= 1220 nm
= 1.22 μm