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VLD [36.1K]
2 years ago
6

According to Newton's thirdlaw, every action has an equal and opposite reaction. When a comb is charged and held near small piec

es of paper, the comb exerts an electric force on the paper pieces and pulls them toward it. Why don't you observe the comb moving toward the paper pieces as well
Physics
1 answer:
NeTakaya2 years ago
5 0

Answer:

Due to the fact that the comb is in terms of mass bigger than pieces of paper

Explanation:

In terms of mass, the comb is way too bigger than the piece of paper so it is virtually impossible for it to move towards the pieces of paper

According to Newton third law action and reaction are equal and opposite

Mathematically

P=m*v

Where

P= impulse is Newton

m=mass of the body

V= velocity of the body

Now since the comb has a greater mass it will not move to the paper, else the paper must have to move to it.

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Justin Bieber is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.
sveticcg [70]

===>  Distance fallen from rest in free fall =

                                         (1/2) (acceleration) (time²)

                               (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²):   (122.5 m / 4.9 m/s²)  =  time²

                                                           (122.5/4.9) s²  =  time²

Take the square root of each side:    5.0 seconds


===> (Accelerating at 9.8 m/s², he will be dropping at
                               (9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'.  We'll need this number for the last part.)


===> With no air resistance, the horizontal component of velocity
doesn't change.

Horizontal distance = (10 m/s) x (5.0 s)  =  50 meters .

===>  Impact velocity =  (10 m/s horizontally) + (49 m/s vertically)

                                 = √(10² + 49²)  =  50.01 m/s  arctan(10/49)

                                 =    50.01 m/s   at  11.5° from straight down,
                                                           
away from the base of the cliff.  

7 0
3 years ago
Read 2 more answers
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
2 years ago
How many electrons are in an electrically neutral atom of boron? A) 3 B) 5 C) 6 D) 8
Hoochie [10]
B. five electrons

If you want an explanation, here it is.

The atomic number gives the number of protons. Protons which have a positive charge are balanced by an equal number of electrons in a neutral atom. Boron number 5 has five protons and therefore as a neutral atom also has five electrons.
8 0
2 years ago
Read 2 more answers
A wave can be described as
rewona [7]
A wave can be described as the disturbance of particles in an area. Think about it this way: particles (matter) carry energy. For all the laws of physics to work, this energy must be "traded" somehow. This happens by miniscule vibrations in the particles, which are apparent disturbances. This creates a wave, and therefore a wave is, indeed, a disturbance.<span />
4 0
3 years ago
When mapping the equipotentials on the plates with different electrode configurations you may find that some have significant ar
Olenka [21]

Answer:

v = 10 V and E = 2 10³ N/C

Explanation:

The electrical potentials and the electric field at one point are related by the expression

            ΔV = - ∫ E. dS

Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.

 If the potential difference is the most usual that is V = 10 V, the electric field is

   s = 0.5 cm = 0.5 10⁻² m

                E = ΔV / S

                E = 10/0.5 10⁻²

                 E = 2 10³ N / C

4 0
3 years ago
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