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.
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Answer:Whenever a moving object experiences friction, some of its kinetic energy is transformed into thermal energy. Mechanical energy is always transformed into thermal energy due to friction. Mechanical energy is always transformed into thermal energy due to friction.
Explanation:
Whenever a moving object experiences friction, some of its kinetic energy is transformed into thermal energy. Mechanical energy is always transformed into thermal energy due to friction. Mechanical energy is always transformed into thermal energy due to friction.
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
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Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
