Answer:
Velocity of both masses after the collisio
Explanation:
Hope it will help
<h2>
<em><u>Brainlists please</u></em></h2>
Answer:
See below explanation
Explanation:
The correspondent chemical reaction for copper carbonate decomposed by heat is:
CuCO₃ (s) → CuO (s) + CO₂ (g)
Considering all molar mass (MM) for each element ( we consider rounded numbers) :
MM CuCO₃ = 123 g/mol
MM CuO = 79 g/mol
MM CO₂ = 44 g/mol
Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.
So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which "escaped" from our system, since it is a gas compound (CO₂)
Answer:
0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.
Explanation:
Answer:
The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.
Explanation:
Answer:
W = 6642 J
Explanation:
Given that,
Mass of a crate, m = 67 kg
Force with which the crate is pulled, F = 738 N
It is moved 9 m across a frictionless floor
We need to find the work done in moving the crate. Let the work done is W. It is given by :
W = F d
W = 738 N × 9 m
= 6642 J
So, the work done is 6642 J.
