Using F=Ma
where a= F/M = 12/3 = 4ms-²
it depends on the atom . but in some cases it may explode
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m 
Where Fe is elastic force, W the weight and the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
Able to conduct electricity
