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GREYUIT [131]
1 year ago
6

A thug pushes some 100 kg punk with 300 N of force. How much is the punk accelerated?

Physics
1 answer:
astra-53 [7]1 year ago
5 0

Answer:

Acceleration generate by punk = 3 m/s²

Explanation:

Given:

Weight of punk = 100 Kg

Force applied on punk = 300 N

Find:

Acceleration generate by punk = ?

Computation:

Acceleration = Force / Mass

Acceleration generate by punk = Force applied on punk / Weight of punk

Acceleration generate by punk = 300 N / 100 Kg

Acceleration generate by punk = 3 m/s²

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The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu
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 maximumforce is F = mg

Explanation:

For this case we must use Newton's second law,

     Σ F = m a

bold indicate vectors, so we will write it in its components x and y

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       Fₓ = maₓ

 Axis y

      Fy - W = m aa_{y}

Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components

      Cos θ = Fₓ / F

      Fₓ = F cos θ

      sin θ = Fy / F

      Fy = F sin θ

Let's replace and calculate

      F sin θ -w = m a

 

As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)

       F sin θ = w = mg

The maximum value of this equation occurs when the sin=1, in this case

      F = mg

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2 years ago
During metamorphism, what is the major effect of chemically active fluids?
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Answer:

Option b

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1 year ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

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tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

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w>v

This means that the centripetal force term

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F=m\frac{w^2}{R}-N sin \theta (3)

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N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
1 year ago
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