<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.
This the answer have a good day
Answer:
I dont know the anwer but i have the same problem
Explanation:
Answer: -
8.00 g
The law of conservation of mass
Explanation: -
When magnesium burns in air it combines with oxygen to form magnesium oxide.
The mass of the product is the mass of the magnesium oxide.
The mass of the reactant is the mass of the magnesium + mass of the oxygen.
Since matter cannot be created or destroyed according to the law of conservation of mass,
Mass of Magnesium oxide formed = mass of magnesium reacted + mass of oxygen reacted
= 3.00 g + 5.00 g
= 8.00 g
Answer:
A 1.0 g sample of propane, C3H8, was burned in the calorimeter.
The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.
Calculate the heat capacity of the calorimeter apparatus in kJ/0C
Explanation:
Given,
The heat of combustion = 10.5kJ/g.
Substitute these values in the above formula to get the value of heat capacity of the calorimeter.
Answer:
The heat capacity of the calorimeter is 
