Weak nuclear force is weaker than the strong nuclear force with a smaller range than the electromagnetic force.
It acts between fermions with spin 1/2 basically quarks and leptons. It has a range of 10⁻¹⁸ meters.
Option 1 is incorrect. It is stronger than the gravitational force.
Option 2 is incorrect. It is weaker than the electromagnetic force.
Option 3 is incorrect. It has a smaller range than the strong nuclear force.
The same thing that holds everything else 'down' --- gravity.
Answer:
ΔV= -5.833×10⁻³
Negative sign indicates that volume decreases
Explanation:
Given data
System heat gains Q=3220 J
Pressure P=1.32×10⁵Pa
Internal energy increases ΔU=3990 J
To find
Change in volume ΔV
Solution
First we need to find the work done
So
W=Q-ΔU
W=3220J-3990J
W= -770J
Now for the change in volume at constant pressure
ΔV=(W/P)
ΔV= -5.833×10⁻³
Negative sign indicates that volume decreases
Answer:
The wavelength of radio waves is 2.85 m.
(B) is correct option.
Explanation:
Given that,
Frequency = 105 MHz
We need to calculate the wavelength of radio waves
Using formula of wave velocity
![v = f\times \lambda](https://tex.z-dn.net/?f=v%20%3D%20f%5Ctimes%20%5Clambda)
![\lambda=\dfrac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bv%7D%7Bf%7D)
Where, v = wave speed
f = frequency
= wavelength
Put the value into the formula
![\lambda=\dfrac{3\times10^{8}}{105\times10^{6}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7B3%5Ctimes10%5E%7B8%7D%7D%7B105%5Ctimes10%5E%7B6%7D%7D)
![\lambda=2.85\ m](https://tex.z-dn.net/?f=%5Clambda%3D2.85%5C%20m)
Hence, The wavelength of radio waves is 2.85 m.
-- 'Ca' (Calcium) is an element.
-- The proton has a positive charge.
-- Nuclear fusion results in the synthesis of atoms of new elements.
-- H₂O (water) is a chemical compound.
-- Nuclear fission is a decay of the nucleus.
-- The atomic number of an element is the number of protons
in each atom of it.
-- I suppose you're using the Greek letter <span>η ('eta', not 'nu')
to represent the neutron.
-- I suppose you're using ' e ' to represent the electron.
</span>