PHYSICS COLLEGE

The surface of an insulating sphere (A) is charged up uniformly with positive charge, and brought very close to an identical–size conducting sphere (B) that has no net charge on it. The spheres do not make contact. A) Sketch the distribution of charge on each sphere.
B) Will the spheres attract, repel, or not interact with each other? Explain.
C) When the spheres make contact, they repel each other. Explain this behavior.

Answers

Answer 1

Answer:

Answer:

A) A negative charge of value Q is induced on sphere B

B) there is an attraction between sphere

C) The charge of sphere A is distributed between the two spheres,

Explanation:

This is an electrostatic problem, in general charges of the same sign attract and repel each other.

with this principle let's analyze the different situations

A) The sphere A that is insulating has a charge on its surface and zero charge is its interior

The conducting sphere B has zero charge, but the sphere A creates an attraction in the electrons, therefore a negative charge of the same value as the charge of the sphere A is induced in the part closest and in the part farther away than one that a positive charge.

A negative charge of value Q is induced on sphere B

B) In this case there is an attraction between sphere A with positive charge and sphere B with negative induced charge

C) When the two spheres come into contact, the charge of sphere A is distributed between the two spheres, therefore each one has a positive charge of value half of the initial charge, as now we have net positive charges in the two spheres charges of the same sign repel each other so the spheres separate

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Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the ball?

Answers

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

Answers

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.

Answers

Answer:

Speed of the wind is 48.989 mph

Explanation:

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind = $(200)/(120+x)$

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind $=(200)/(120-x)$

Total time is given as t= 4 hour

So $(200)/(120+x)+(200)/(120-x)=4$

$(200(120-x)+200(120+x))/((120+x)(120-x))=4$

$48000=57600-4x^2$

$4x^2=9600$

x = 48.989 mph

Answer:

Explanation:

Type Distance Rate Time

Headwind 200 120-r 200/120-r

Tailwind 200 120 - r 200/120 - r

We know the times add to 4, so we write the equation:

200/120−r + 200/120 + r = 4

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

Learn more about energy of photons here:

brainly.com/question/34240307

#SPJ3

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 T and direction from east to west.A)If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.

Use 1.602×10−19 C for the magnitude of the charge on an electron.

B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.

a-from north to south

b-from south to north

Answers

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

$v_(f)^(2)=v_(0)^(2)+2gy\nv_(f)=\sqrt{0+2(9.8(m)/(s^(2)))(145m)}=53.31(m)/(s)$

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

$F=(3.8*10^(8))(1.602*10^(-19)C)(53.31(m)/(s))(0.205T)=6.65*10^(-10)N$

b. south - north (by the rigth hand rule)

I hope this is usefull for you

regards

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?a) TA > TB
b) TA < TB
c) TA = TB
d) More information is needed

Answers

The final temperatures are such that TA > TB.

The specific heat capacity refers to the quantity of heat required to raise the temperature of 1 Kg of a body by 1K. The higher the specific heat capacity of a body, the higher the quantity of heat required to raise the temperature of the body and vice versa.

Hence, if the specific heat of substance A is greater than that of substance B and A and B are at the same initial temperature, when equal amounts of energy are added to them, the final temperature are such that TA > TB.

Learn more: brainly.com/question/1445383

Answer:

$m_A c_(pA) (T_(fA) -T) = m_B c_(pB) (T_(fB)- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_(pA) > c_(pB)$