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GuDViN [60]
3 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]3 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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A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same mome
lozanna [386]

A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same momentum?A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same momentum?

8 0
3 years ago
A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t
irga5000 [103]

Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

6 0
3 years ago
Answer choices
marshall27 [118]

Answer:

Option-C (Lipoprotein profile)

4 0
2 years ago
If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?​
Marina CMI [18]
The acceleration rate would be .14667 m/s^2
6 0
3 years ago
A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
USPshnik [31]

Answer:

Temperature increase = 2.1 [C]

Explanation:

We need to identify the initial data of the problem.

v = velocity of the copper sphere = 40 [m/s]

Cp = heat capacity = 387 [J/kg*C]

The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:

E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT  \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]

8 0
3 years ago
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