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Wittaler [7]
3 years ago
15

Consider a situation where you are playing air hockey with a friend. The table shoots small streams of air upward to keep the pu

ck afloat and to minimize friction. When you barely tap the puck forward, it
Question 4 options:
continues to accelerate.
moves with a constant speed until hitting the other end.
moves backward with an equal and opposite velocity.
slows and stops.
Physics
1 answer:
artcher [175]3 years ago
5 0

When we hit the puck from tap the puck will move forward.

This is due to the impulse provided by us at the time of hit. Due to this impulse the puck will move forward and start moving in some direction.

As soon as puck move forward the force on it is zero as the weight of the puck is counterbalanced by the air stream force and there is no other force on it so puck will continue its motion till it will hit at some other point.

So here the motion of the puck will be uniform motion till it will collide with some other points.

So here the correct option will be given as

<em>moves with a constant speed until hitting the other end.</em>

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Of the two forms of friction you studied,_____ friction is the one that must be overcome in order to start something moving.​
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How much energy is stored in the electric field of a 50-μm-diameter cell with a 7.0-nm-thick cell wall whose dielectric constant
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A fairgrounds ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo
Valentin [98]

We know that the acceleration due to gravity g is: g = 9.81 m/s^2

So the centripetal acceleration (w) is:

w^2 = 1.5 g / r

w^2 = 1.5 * (9.81 m/s^2) / 5 m

w = 1.716 rad / s

To convert to rad to rev:

w = (1.716 rad / s) * (1 rev / 2π rad) * (60 s/min)

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3 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
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