When the tuning forks are struck simultaneously, the beat frequency is equal to the absolute value of the difference between the frequencies of the two tuning forks:
![f_b = |f_1 - f_2|](https://tex.z-dn.net/?f=f_b%20%3D%20%7Cf_1%20-%20f_2%7C)
where
![f_b](https://tex.z-dn.net/?f=f_b)
is the beat frequency, and
![f_1, f_2](https://tex.z-dn.net/?f=f_1%2C%20f_2)
are the frequencies of the two forks.
In the first part of the problem,
![f_1 = 240 Hz](https://tex.z-dn.net/?f=f_1%20%3D%20240%20Hz)
and
![f_b=4 Hz](https://tex.z-dn.net/?f=f_b%3D4%20Hz)
, with
![f_2](https://tex.z-dn.net/?f=f_2)
being the frequency of the unknown fork. In the second part of the problem,
![f_1=250 Hz](https://tex.z-dn.net/?f=f_1%3D250%20Hz)
and
![f_2 = 6 Hz](https://tex.z-dn.net/?f=f_2%20%3D%206%20Hz)
. So we have the following system of two equations:
![4 Hz = |240 Hz -f_2|](https://tex.z-dn.net/?f=4%20Hz%20%3D%20%7C240%20Hz%20-f_2%7C)
![6 Hz = |250 Hz - f_2|](https://tex.z-dn.net/?f=6%20Hz%20%3D%20%7C250%20Hz%20-%20f_2%7C)
and the solution of this system is
Answer:
A nucleotide consists of a sugar molecule (either ribose in RNA or deoxyribose in DNA) attached to a phosphate group and a nitrogen-containing base. The bases used in DNA are adenine (A), cytosine (C), guanine (G), and thymine (T).
Answer:
The answer should be Playing Area
Explanation:
I HOPE THIS HELPS.....GOOD LUCK!!!
Answer:
n=142 N
Explanation:
solution:
pic 1 is attached
<em>There is no vertical acceleration (ay = 0), so net sum of vertical forces is 0. n acts up, weight and the vertical component of F act down, so </em>
∑F_y =m*a_y
n-mg-Fsin43°=0
n=142 N
Answer:
18375g
Explanation:
![\boxed{density = \frac{mass}{volume} }](https://tex.z-dn.net/?f=%5Cboxed%7Bdensity%20%3D%20%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%7D)
![∴ mass = density \times volume](https://tex.z-dn.net/?f=%E2%88%B4%20mass%20%3D%20density%20%5Ctimes%20volume)
Let's find the volume of the rectangular block.
Volume
= length ×breadth ×height
= 3.5 ×6 ×2.5
= 52.5cm³
Mass of the block
= 350(52.5)
= 18375g