Answer:
The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.
(3) and (4) is correct option.
Explanation:
Given that,
Beat frequency f = 5.0 Hz
Frequency f'= 462 Hz
We need to calculate the possible frequencies for the A string of the other violinist
Using formula of frequency
...(I)
...(II)
Where, f= beat frequency
f₁ = frequency
Put the value in both equations
Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.
Answer:
The current value of the Hubble's constant = 73 km/sec/Mpc.
t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc
Explanation:
The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.
Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.
Hence,
v is directly proportional to d
where, v = apparent velocity
d = distance
if we equate velocity and distance then there comes Hubble constant.
v = x d
= 73 km/sec/Mpc
where, Mpc = Mega Parsec = 1 Mpc = 3.086 x km
We can use Hubble constant to tell the age of universe.
t = d/v
t = d/( xd)
t = 1/
Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.
Now, if we hypothetically change the value of Hubble constant,
from = 73 km/sec/Mpc to
= 700 km/sec/Mpc
then the age of universe will be:
t = 1/
first convert the units of new into 1/s
= (700) x (/3.08 x
)
= 227.27 x
= 2.27 x
1/s
So,
Age of universe will be:
t = 1/ = 1/2.27x
1/s
t = 2.27 x s
t = 71.9 trillion years
t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc