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
<span>experimental? I hope that helps :)</span>
Answer:
I do belive that it is B hrs cn I an gn
Answer:
A
Explanation:
Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.
Lots of reasons. one reason i lie alot (a very bad habit) is im scared of what will happen if i tell the truth. the truth is always better, though.
