<h3>
Answer:</h3>
83.33 seconds.
<h3>
Explanation:</h3>
<u>We are given;</u>
- Take off velocity as 300 km/hr
- Acceleration as 1 m/s²
We are required to calculate the take off time of the airplane.
<h3>Step 1: Convert velocity from km/hr to m/s </h3>
We are going to use the conversion factor.
The conversion factor is 3.6 km/hr per m/s
Therefore;
Velocity = 300 km/hr ÷ 3.6 km/hr per m/s
= 83.33 m/s
<h3>Step 2: Calculate the take off time</h3>
We know that;
v = u + at
where, u is the initial velocity, v the final velocity, a the acceleration and t is time.
But, initial velocity is Zero
Therefore;
83.33 m/s = 1 m/s² × t
Thus;
time = 83.33 m/s ÷ 1 m/s²
= 83.33 seconds
Therefore, the take off time is 83.33 seconds.
Answer:
2–butyne.
Explanation:
To name the compound given above, we must determine the following:
1. Determine the functional group of the compound.
2. Determine the longest continuous carbon chain. This gives the parent name of the compound.
3. Locate the position of the functional group by giving it the lowest possible count.
4. Combine the above to obtain the name.
Thus, we shall name the compound as follow:
1. The compound contains triple bond (C≡C). Therefore, the compound is an alkyne.
2. The longest chain is carbon 4. Thus the parent is butyne.
3. The triple bond (C≡C) is located at carbon 2 when we count from either side.
4. The name of the compound is:
2–butyne
Hello there,
Ione pairs of electrons that are present in a molecule of C2O are...
4
Each Oxygen forms two bonds with Carbon
Hope I Helped!
-Char
