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The minimum charge on any object cannot be less than​

Answers

Answer 1
Answer:

Answer:

1.6 x 10^{-19} Coulombs

Explanation:

In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

The minimum charge on any object such as an electron cannot be less than​ 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.

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A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

E_i=(mv^2)/(2)+mgh

Here , initial velocity is zero .

Therefore , E_i=mgh=40\ J

Now , final energy of the system :

E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

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5. A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with the velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

Answers

Answer:

The recoil velocity of the raft is 1.205 m/s.

Explanation:

given that,

Mass of the swimmer, m_1=55\ kg

Mass of the raft, m_2=210\ kg

Velocity of the swimmer, v = +4.6 m/s

It is mentioned that the swimmer then runs off the raft, the total linear momentum of the  swimmer/raft system is conserved. Let V is the recoil velocity of the raft.

m_1v+m_2V=0

55* 4.6+210V=0

V = -1.205 m/s

So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.

Answer:

The recoil velocity of the raft would be v_(r)\approx 1.2(m)/(s) (pointing to the left if the swimmer runs to the right)

Explanation:

The problem states thatthe swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.

To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.

Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:

p_(i)=0

Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as

p_(f)=mv-Mv_(r)=0

The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.

Then, from that previous relation we can clear

v_(r)=(m)/(M)v=(55)/(210)*4.6(m)/(s)=(253)/(210)(m)/(s)\approx1.2(m)/(s)

wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

Tan \theta = (y)/(D)

where  D  is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy  =  |y1max  (λ1) −  y1max  (λ2)|

Δy = |(D\lambda _1)/(d) - (D\lambda _2)/(d) |

Δy = D |(d/20)/(d) - (d/15)/(d) |

Δy = D |(1)/(20) - (1)/(15) |

Δy = 4.90 |(1)/(20)- (1)/(15) |

Δy = 81.7mm

The separation between these maxima is 81.7 mm

b)

Δy  =  |y₂max  (λ1) −  y₂max  (λ2)|

Δy = D|(2(d/20))/(d) - (5(d/15))/(2d) |

Δy = 4.90|(1)/(10) - (1)/(6) |

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

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State the following forms of electromagnetic radiation in increasing order of wavelength.Radiowaves, gamma rays, x-rays, infrared radiation, visible light​

Answers

Answer:

Gamma rays, x-rays, visible light, infrared radiation and radiowaves

Explanation:

Gamma rays, x-rays, ultraviolet, visible light, infrared radiation, microwave and radiowaves

Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?

Answers

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

One of two nonconducting spherical shells of radius a carries a charge Q uniformly distributed over its surface, the other carries a charge -Q, also uniformly distributed. The spheres are brought together until they touch.A.) What does the electric field look like, both outside and inside the shells?

B.) How much work is needed to move them far apart?

Answers

Answer:

Explanation:

A ) The spheres are non conducting , charge will not move on the surface so neutralization of charge by + ve and - ve charge is not possible. Charges will remain intact on them . The electric field inside them will be zero . Electric field outside shell  will not be spherically symmetrical . Lines of force will emanate from the surface of positively charged shell  outwardly oriented  and end at negatively charged shell .

B )

distance between the centres of spherical shell

= 2 a

potential energy of charges

= k q₁ x q₂ / R

= k x - Q x Q / ( 2a )

= - k Q²/ 2a

So work needed to separate them to infinity will be equal to

= k Q²/ 2a

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