Answer:
The answer is below
Explanation:
Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum with respect to time, going in the same direction as the force.
Let F = force, m = mass of object, v = velocity of object, mv = momentum.
F = d/dt(mv) = m(dv / dt) = ma; a = acceleration.
Let us assume that the object starts from rest to 5 m/s within 1 seconds, hence:
F = m(dv / dt)
200 N = m[(5 m/s - 0 m/s) / (1 s)]
200 = 5m
m = 40 kg
Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:

= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
First let us see what
kind of bonds are formed in the compound. By drawing the structure, we see that
the kind of bonds are:
N =- triple bond -= C –
O
<span>So there is only
single bond between C and O therefore the hybridization of C is sp.</span>
Double replacement because H and K are both switching
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