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GuDViN [60]
4 years ago
11

A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the

electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Physics
1 answer:
Anton [14]4 years ago
6 0

Answer:

1.176\times 10^{-3} N

Upward

Explanation:

We are given that

Mass of scarp paper,m=120 mg=120\times 10^{-6}kg

1mg=10^{-6} mkg

Distance,d =8 mm=8\times 10^{-3} m

Magnitude of electric force =F_E= w=mg

Where g=9.8 m/s^2

Substitute the values

F_E=120\times 10^{-6}\times 9.8

F_E=1.176\times 10^{-3} N

Gravitational force act in  downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

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Answer:

0.198 s

Explanation:

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Using the kinematics equation

v_{f} = v_{o} + at \\v_{f} = 0 + (-g)t\\v_{f} = - gt

m = mass of the bullet = 0.026 kg

v_{b} = velocity of block just before collision = 750 m/s

M = mass of the block = 5 kg

V = final velocity of bullet block after collision = gt

Using conservation of momentum

m v_{b} + Mv_{f} = (m + M) V\\(0.026) (750) + (5) (- gt) = (0.026 + 5) (gt)\\19.5 - 49 t = 49.2548 t\\t = 0.198 s

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3 years ago
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5 0
3 years ago
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The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,
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Answer:

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Explanation:

The information that we have is

Intensity at 2.0 m: I=80dB and r_{1}=2m

we need an intensity level of: I_{2}=40dB

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we solve for r_{2}:

\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }

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