Answer:
Explanation:
m1 = 24 kg
m2 = 63 kg
(a) Let the tension is T1 in upper rope and T2 in lower rope.
Use Newtons' second law on m2.
T2 = 856.8 N
Apply Newton's second law on m1
T1 = 1183.2 N
(b)
Use Newtons' second law on m2.
T2 = 378 N
Apply Newton's second law on m1
T1 = 522 N
It's always a good idea to start with the definition of the thing you're trying to find.
This problem is just trying to find out whether you KNOW the definition of acceleration. You may know it, but you haven't used it yet.
Average acceleration = (change in velocity) divided by (total time).
Change in velocity = (end value) minus (start value)
Change in velocity = (20m/s) - 0
Change in velocity = 20 m/s
Time = 10 s
Average acceleration = (20m/s)/(10s)
Average acceleration = 2 m/s^2
Hey there,
It is C, Platelets
There are 4 types of macromolecules. They are Carbohydrates, Lipids, Proteins and Nucleic Acids.
Hope this helps :))
~Top
Answer:
Explanation:
The reaction between hypothetical elements X and Y to produce hypothetical compound XY₂ is represented with the chemical equation:
That equation means that one atom (or one mole of atoms) of element X reacts with 2 atoms (or two moles of atoms) of element Y to produce one molecule (one mole of molecules) of compound XY₂.
Elements X and Y are the reactants, and compound XY₂ is the product.
Hence, the mole ratio is:
- 1 mol X : 2 mol Y : 1 mol XY₂
Since, 10.0 mol of Y completely react you can set the corresponding proportions with X and XY₂ to find how many moles of them will be in the reaction:
<u>1. Moles of X</u>:
- Theoretical ratio: 1 mol of X / 2 mol of Y
- Ratio with 10.0 mol of Y: U / 10.0 mol of Y
- Proportion: equal the two ratios: 1 mol X / 2 mol Y = U / 10.0 mol Y
- Solve for U: U = 10.0 mol Y × 1 mol X / 2 mol Y = 5.0 mol X
- Result: 5.0 moles of X are in the reaction.
<u>2. Total moles of X and Y in the reaction:</u>
- The total number of moles is the sum of the moles of the two reactants: 10.0 moles of Y + 5.0 moles of X = 15.0 moles. Hence 15.0 mol of X and Y are in the reaction: option C).