Question

A helium atom has a mass of 6.645 X 2 points
10^-27 Kg. When attached to a
certain massive molecule it oscillates
with a frequency of 10^9 cycles per
second with an amplitude of 10^-6
m. Calculate the force acting on the
Helium atom.
a= 39.43x10^11 m/s^2, F = 2.62x10^-12 N
O a= 39.43*10^12 m/s^2, F = 2.62x10^-13 N
O a/^
a= 39.43x10^13 m/s^2, F = 2.62x10^-14 N
a= 30.43x10^12 m/s^2, F = 2.0x10^-13 N
a= 29.43x10^12 m/s^2, F = 1.62x10^-13 N​

Answer 1

Answer:c

Explanation:

Answer 2

The force acting on Helium atom is  2.620 x $10^{-13}$N

Given:

helium atom mass =6.645 x $10^{-27}$

frequency = 10^9 cycles per second

amplitude = 10^-6 m

To find:

the force acting on the helium atom.

solution:

we know that,

ω² = k/m

(2πf)² = k/m

k = 4π²f²m and,

F = $kx_{0}$

  = 4π²f²m$x_{0}$  

F = 4 x (3.14)² x $10^{18}$ x 6.645 x $10^{-27}$ x $10^{-6}$

F = 2.620 x $10^{-13}$N

hence,  force acting on the Helium atom is  2.620 x $10^{-13}$N

​

​