Average speed = total distance / total time
total distance = 40 + 20 = 60km
total time taken = 10 + 5 = 15 minutes
Average speed = 60/15 = 4km/min
The various contributions involved till the chapati is made is given below.
<h3>What is food?</h3>
The substance that we intake for the body to charge up by giving nutrients is called the food.
Wheat is a staple food. We make chapati from flour obtained from the wheat grains.
The various contributions involved till the chapati is made is given below.
Take required amount of atta in a container
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Add water accordingly to form a dough
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Apply oil to make dough smooth for long time
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Take small dough, make it a ball shaped and apply dry flour
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Roll it using rolling pin on the chapati maker plate
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After making it circular or any shape you want, place it on hot tawa
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Bake it on both the sides
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Chapati is ready
Thus, the flow chart is made.
Learn more about food.
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Heat required to raise the temperature of a given system is

here we know that
m = mass
s = specific heat capacity
= change in temperature
now as we know that
mass of wood = 5 kg
mass of aluminium pan = 2 kg
change in temperature = 45 - 20 = 25 degree C
specific heat capacity of wood = 1700 J/kg C
specific heat capacity of aluminium = 900 J/kg C
now here we will find the total heat to raise the temperature of both




So heat required to raise the temperature of the system is 257500 J
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
(1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:

Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:

Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
Answer:

Explanation:
Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:




Ratio of kilomoles oxygen to kilomole nitrogen is:


It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.
The empirical formula for the compound is:
