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3 years ago

1. How are each of Newton’s Laws of motion demonstrated

Physics

1 answer:

stepan [7]3 years ago

5 0

Answer:

Newton's law of inertia - His first law states that every object remains at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... This is the first part cited in Newton's first law; "there is no net force on the airplane and it travels at a constant velocity in a straight line."

Newton's law of acceleration - "a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity."

Newton's law of Action/Reaction - "As a plane flies, the force of the air hitting the plane is always equal and opposite to the force of the plane pushing against the air. The force generated by the engine pushes against air while the air pushes back with an equal and opposite force."

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monitta

The various contributions involved till the chapati is made is given below.

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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6 0

2 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

5 0

3 years ago

A commonly used unit of electrical energy.

ladessa [460]

The two most common units of electric energy is Watts or hertz.

8 0

3 years ago

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin

JulsSmile [24]

Explanation:

(a) Since, it is given that the blocks are identical so distribution of charge will be uniform on both the blocks.

Hence, final charge on block A will be calculated as follows.

Charge on block A = $\frac{(8.7 + 0 nC}{2}$

= 4.35 nC

Therefore, final charge on the block A is 4.35 nC.

(b) As it is given that the positive charge is coming on block A . This means that movement of electrons will be from A to B.

Thus, we can conclude that while the blocks were in contact with each other then electrons will flow from A to B.

6 0

3 years ago

Read 2 more answers

A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.

Ainat [17]

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

$\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg$