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3 years ago

Newton’s third law of motion is also known as action/_______?

2 answers:

8 0

Newton's third law<span> is: For every </span>action, there is an equal and opposite reaction. This means that in every interaction, there is a<span> pair of forces acting on the two interacting objects.</span>

wel3 years ago

7 0

Newton's third law of motion is also known as action/reaction

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A feris wheel is turning at a constant speed of 5m/s is it not accelerating ,true or false

erma4kov [3.2K]

Answer:no it is staying the same speed

Explanation:

6 0

3 years ago

Identify the arrows that show the movement of carbon from the biosphere to the atmosphere in the carbon cycle.

Illusion [34]

Answer:

All of the arrows pointing up that have a red box NOT the arrow pointing down with a red box. (if the blue squigglies on the water are arrows then they count too, the picture is not too clear)

Explanation:

8 0

3 years ago

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke

umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

Thrust = $v_{e} \frac{dM}{dt}$

$v_{f}$ -v₀ = v_{e} $ln ( \frac{M_{o} }{M_{f}} )$

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

The final mass is the mass of the engines + the mass of the rocket

M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

thrust = 5.26 N

t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

thrust = v_{e} $\frac{M_{f} - M_{o} }{t_{f} - t_{o} }$

v_{e} = thrust $\frac{\Delta t}{\Delta M}$

v_{e} = 5.26 $\frac{1.90}{0.080 -0.0927}$

v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

v_{f}-0 = v_{e} $ln ( \frac{M_{o} }{M_{f} } )$

we calculate

v_{f} = 786.93 ln (0.0927 / 0.080)

v_{f} = 115.95 m / s

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3 years ago

If your brother changes the channel, you will change it back, so then he will change it, then you will change it, etc, etc. This

KonstantinChe [14]

The answer is the law of inetia

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3 years ago

URGENT!!

irinina [24]

Answer:

Explanation:La ecuación de Van der Waals es una ecuación de estado de un fluido compuesto de partículas con un tamaño no despreciable y con fuerzas intermoleculares, como las fuerzas de Van der Waals. La ecuación, cuyo origen se remonta a 1873, debe su nombre a Johannes van der Waals, quien recibió el premio Nobel en 1910 por su trabajo en la ecuación de estado para gases y líquidos, la cual está basada en una modificación de la ley de los gases ideales para que se aproxime de manera más precisa al comportamiento de los gases reales al tener en cuenta su tamaño no nulo y la atracción entre sus partículas.

3 0

3 years ago