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Flura [38]
2 years ago
12

A negatively charged plastic comb is brought close to, but does not touch, a small piece of paper. If the comb and the paper are

attracted to each other the charge on the paper
A) must be negative
B) may be positive or neutral
C) may be negative or neutral D)must be positive
Physics
1 answer:
Irina-Kira [14]2 years ago
4 0

If the comb and the paper are attracted to each other the charge on the paper, is D)must be positive

<h3>Laws of electrical attraction</h3>

This states that

  • Like charges attract
  • Unlike charges repel

Now, given that a negatively charged plastic comb is brought close to, but does not touch, a small piece of paper. If the comb and the paper are attracted to each other the charge on the paper, this implies that both the negatively charged plastic comb and the paper have opposite charges.

Since the charge on the plastic comb is negative, this means that the charge on the paper must be positive

So, if the comb and the paper are attracted to each other the charge on the paper, is D)must be positive

Learn more about electric charge here:

brainly.com/question/2373424

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A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
Explain how energy is transferred in an impact situation such as a car crash.
DaniilM [7]
During a car crash, energy is transferred from the vehicle to whatever it hits, be it another vehicle or a stationary object. ... The object that was struck will either absorb the energy thrust upon it or possibly transfer that energy back to the vehicle that struck it.

I HOPE THIS HELPSS???
Mark me brainliest
8 0
3 years ago
How fast must a bug swim to keep up with the waves it produces? How fast must it move to produce a bow wave?
Elden [556K]

Answer:

A bug must swim as fast as the wave speed to keep up with the waves it produces. Moreso, a boat must be moving faster than the waves it creates to produce a bow wave.

6 0
4 years ago
25 POINTS FOR CORRECT ANSWER
castortr0y [4]
No.  I do not agree with Stefan.  Quite the contrary.  I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray. 

The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray. 

Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>
5 0
3 years ago
Read 2 more answers
A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)
kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = \frac{4\pi}{3}\times(0.0991^2-0.0882^2)

or

volume of sphere = 0.0012 m³

also, volume of half sphere = \frac{\textup{Total volume}}{\textup{2}}

or

volume of half sphere = \frac{\textup{0.0012}}{\textup{2}}

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density =  \frac{\textup{Mass}}{\textup{Volume}}

or

Density = \frac{\textup{0.568}}{\textup{0.0012}}

or

Density = 474 kg/m³

8 0
3 years ago
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