Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
Answer:
Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell
<span>A: It is not an exact representation of the atom, but is close enough to be very useful.
Hope this helps!</span>
