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rewona [7]
2 years ago
9

Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th

e following questions.
(a) What new force will exist if qA is doubled?
(b) What new force will exist if qA and qB are cut in half?
(c) What new force will exist if d is tripled?
(d) What new force will exist if d is cut in half?
(e) What new force will exist if qA is tripled and d is doubled?
Physics
1 answer:
maria [59]2 years ago
6 0

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

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An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the
Nataly [62]

Answer:

The mass of the object is 49.5kg which is approximately 50kg

Explanation:

Given that

Spring constant (k)=45N/m

The extension (e)=0.88m

Also given that the acceleration is 0.8m/s²

Force by the spring is given as

Using hooke's law

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

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k is the spring constant

e is the extension

From the formula k = F/e

F=ke

m is the mass of the block = ?

a is the acceleration = 0.8m/s²

e is the extension of the spring =  0.88m

k is the spring constant = 45N/m

F=45×0.88

F=39.6N

Now this force will set the object in motion, now using newton second law of motion

F=ma

Then, m=F/a

m=39.6/0.8

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The mass of the object is 49.5kg which is approximately 50kg

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To ensure the greatest nutritional benefits, dietary guidelines suggest which of the following?
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The answer would be D. Eating nutrient-dense foods
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Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. If titanium is irradiated with light of
butalik [34]

Answer:

a) 1.59(10)^{-19} J

b) 2.34(10)^{12} electrons

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.  

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy  E:

E=\frac{hc}{\lambda} (1)  

So, the energy E of the incident photon must be equal to the sum of the Work function \Phi of the metal and the kinetic energy K of the photoelectron:  

E=\Phi+K (2)  

Where \Phi=6.94(10)^{-19} J is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a)  Maximum possible kinetic energy of the emitted electrons (K)</h3>

From (1) we can know the energy of one photon of 233 nm light:

E=\frac{hc}{\lambda}

Where:

h=6.63(10)^{-34}J.s is the Planck constant  

\lambda=233 (10)^{-9} m is the wavelength

c=3 (10)^{8} m/s is the speed of light

E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s} (3)

E=8.53(10)^{-19} J (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

8.53(10)^{-19} J=6.94(10)^{-19} J+K (5)  

Finding K:

K=1.59(10)^{-19} J (5)  This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is 2 \mu J=2(10)^{-6} J</h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

\frac{E_{burst}}{E}

Where E_{burst}=2(10)^{-6} J is the energy of the burst of light

Hence:

\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons This is the maximum number of electrons that can be freed by the burst of light.

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A wheel with rotational inertia 0.04 kg•m2 and radius 0.02 m is turning at the rate of 10 revolutions per second when a friction
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Answer:

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