You could use grams hope this helps
Rust is a chemical all the others are physical changes
Answer:
2.35 m/s²
Explanation:
Given that
Mass of the smaller crate, m₁ = 21 kg
Mass of the larger crate, m₂ = 90 kg
Tensión of the rope, T = 261 N
We know that the sum of all forces for the two objects with a force of friction F and a tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.
1) no sliding can also mean that:
a₁ = a₂ = a
This makes us merge the two equations written above together as:
m₂a = T - m₁a
If we then solve for a, we would have something like this
a = T / (m₁+m₂)
a = 261 / (21 + 90)
a = 261 / 111
a = 2.35 m/s²
Therefore, the needed acceleration of the small crate is 2.35 m/s²
Answer:
the final velocity of the two blocks is 
the distance that A slides relative to B is 
Explanation:
From the diagram below;
acceleration of A relative to B is : 
where
v = u + at
Making t the subject of the formula; we have:
which implies the distance that A slides relative to B.
The final velocities of the two blocks can be determined as follows:
v = u + at
Thus, the final velocity of the two blocks is
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>
