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-BARSIC- [3]
3 years ago
15

A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.

Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

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Researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the fil
earnstyle [38]

Answer:

Minimum thickness; t = 9.75 x 10^(-8) m

Explanation:

We are given;

Wavelength of light;λ = 585 nm = 585 x 10^(-9)m

Refractive index of benzene;n = 1.5

Now, let's calculate the wavelength of the film;

Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene

Thus; λ_film = 585 x 10^(-9)/1.5

λ_film = 39 x 10^(-8) m

Now, to find the thickness, we'll use the formula;

2t = ½m(λ_film)

Where;

t is the thickness of the film

m is an integer which we will take as 1

Thus;

2t = ½ x 1 x 39 x 10^(-8)

2t = 19.5 x 10^(-8)

Divide both sides by 2 to give;

t = 9.75 x 10^(-8) m

8 0
3 years ago
Do true blackbodies exist in nature?<br><br> Yes or No??
MaRussiya [10]

Answer:

no

Explanation:

4 0
3 years ago
Read 2 more answers
What is the potential energy of a 3kg ball that is on the ground?
ELEN [110]

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
'relative to what?'.

         Potential energy = (mass) x (gravity) x (<em>height</em>).

So you have to specify <em><u>height above what</u></em> .

-- With respect to the ground, the ball has zero potential energy.
(If you let go of it, it will gain zero kinetic energy as it falls to
the ground.)

-- With respect to the floor in your basement, the potential energy is

                 (3) x (9.8) x (3 meters) = 88.2 joules.

(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
to the floor of your basement.)

-- With respect to the top of that 10-meter hill over there, the potential
energy is
                    (3) x (9.8) x (-10) = -294 joules

(Its potential energy is negative. After you let go of it, you have to give it
294 joules of energy that it doesn't have now, in order to lift it to the top of
the hill <em>where it will have zero</em> potential energy.)


5 0
3 years ago
QUESTION 1
mario62 [17]
It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.
7 0
3 years ago
Read 2 more answers
The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

7 0
3 years ago
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