<h3>
Answer:</h3>
The centripetal acceleration is 26.38 m/s²
<h3>
Explanation:</h3>
We are given;
- Mass of rubber stopper = 13 g
- Length of the string(radius) = 0.93 m
- Time for one revolution = 1.18 seconds
We are required to calculate the centripetal acceleration.
To get the centripetal acceleration is given by the formula;
Centripetal acc = V²/r
Where, V is the velocity and r is the radius.
Since time for 1 revolution is 1.18 seconds,
Then, V = 2πr/t, taking π to be 3.142 ( 1 revolution = 2πr)
Therefore;
Velocity = (2 × 3.142 × 0.93 m) ÷ 1.18 sec
= 4.953 m/s
Thus;
Centripetal acceleration = (4.953 m/s)² ÷ 0.93 m
= 26.38 m/s²
Hence, the centripetal acceleration is 26.38 m/s²
Answer : The mass of copper(I) oxide is, 1180.5 grams.
Explanation : Given,
Moles of copper(I) oxide = 8.250 mol
Molar mass of copper(I) oxide = 143.09 g/mol
Formula used:
Now put all the given values in this formula, we get:
Therefore, the mass of copper(I) oxide is, 1180.5 grams.
Answer:
1.5 litre of C5H12 produce 9 litre of water vapour;
Explanation:
First balance the combustion reaction;
→ 
from the balance reacion it is clearly that,
one mole of C5H12 reacts completely with 8 mole of O2 gas;
Hence
one litre of C5H12 reacts completely with 8 litre of O2 gas;
there fore 1.5 litre C5H12 needs 12 litre oxygen gas but we have 15 litre.
so,
C5H12 is a limiing reactant and O2 is a excess reactant.
so quantity of H2O depends on limiting reactant;
one litre of C5H12 produce 6 litre of water vapour;
therefore,
1.5 litre of C5H12 produce 9 litre of water vapour;
The jet stream is a band of reliably strong wind that plays a key role in keeping colder air north and warmer air south. But when the vortex weakens, part of the weakened low-pressure system can break off. This breaking-off process is what causes a polar vortex.
Without that strong low-pressure system, the jet stream does not have enough force to maintain its usual path. It becomes wavy and rambling. When high-pressure systems get in its way, a collection of cold air pushes south, along with the rest of the polar vortex system
