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12345 [234]
3 years ago
11

When discussing Newton’s laws of motion, which terms do people most likely use when talking about Newton’s third law of motion?

"force" and "acceleration" "inertia" and "force" "action" and "reaction" "mass" and "inertia"
Physics
2 answers:
mezya [45]3 years ago
6 0

Answer:

INERTIA

Explanation:

KatRina [158]3 years ago
4 0

Answer:

Action and reaction

Explanation:

Newton's third law states that for every action there is an equal and an opposite reaction. The pair of force exits there. The pair of force is "action and reaction".

For example, if you slap someone, then you also feel pain in your hand. If you slap someone then this is an action. You feel pain in your hand then it is a reaction.

Force and acceleration terms are related to Newton's second law of motion.

The relation between force and acceleration is as;        

F = ma

Here, m is the mass, F is the applied force and a is the acceleration.

Inertia and force terms are related to Newton's first law of motion.  Newton's first law of motion is also known as the law of inertia. Inertia is a tendency to remain in a state unchanged.

Therefore, the "action and reaction" terms are related to Newton's third of motion.

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A ball is released at the top of a ramp at t = 0. Which is the speed of the ball at t = 4?
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his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi
zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate\\\rho = Density\\V = Velocity

values are divided by inlet(1) and outlet(2) by

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A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s

PART A) Applying the flow equation

\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2

7 0
3 years ago
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