PHYSICS COLLEGE

A research Van de Graaff generator has a 2.00-m diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface?
(b) At what distance from its center is the potential 1.00 MV?
(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?

Answers

Answer 1

Answer:

Answer:

a) $V=49.5MV$

b) $r=49.5m$

c) $\Delta U=3*(49.5-1)=145.5 MeV$

Explanation:

a) The potential equation is given by:

$V=k(Q)/(r)$

k is the electrostatic constant ( $k=9.9*10^(9)Nm^(2)/C^(2)$ )

Q is the charge Q = 5mC

r is the radius of the sphere r = 1 m

$V=9.9*10^(9)(5*10^(-3))/(1)=49.5MV$

b) We solve it using the same equation.

Here we need to find r:

$r=k(Q)/(V)$

$r=9.9*10^(9)(5*10^(-3))/(1*10^(6))$

$r=49.5m$

c) The relation between difference potential and electrical energy is:

$\Delta U=\Delta Vq$

here q is 3e becuase oxygen atom has three missing electrons

Therefore:

$\Delta U=3*(49.5-1)=145.5 MeV$

I hope it heps you!

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A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Answers

Answer:

a) The initial speed of the rock is approximately 14.607 meters per second.

b) The greatest height of the rock from the base of the cliff is 42.878 meters.

Explanation:

a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:

$U_(g,1)+K_(1) = U_(g,2)+K_(2)$ (Eq. 1)

Where:

$U_(g,1)$ , $U_(g,2)$ - Initial and final gravitational potential energies, measured in joules.

$K_(1)$ , $K_(2)$ - Initial and final translational kinetic energies, measured in joules.

By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:

$m\cdot g\cdot (y_(1)-y_(2))= (1)/(2)\cdot m\cdot (v_(2)^(2)-v_(1)^(2))$

$g\cdot (y_(1)-y_(2)) = (1)/(2)\cdot (v_(2)^(2)-v_(1)^(2))$ (Eq. 2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$y_(1)$ , $y_(2)$ - Initial and final height, measured in meters.

$v_(1)$ , $v_(2)$ - Initial and final speed of the rock, measured in meters per second.

If we know that $g = 9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $y_(2) = 0\,m$ and $v_(2) = 29\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-0\,m) = (1)/(2)\cdot \left[\left(29\,(m)/(s) \right)^(2)-v_(1)^(2)\right]$

$313.824 = 420.5-0.5\cdot v_(1)^(2)$

$0.5\cdot v_(1)^(2) = 106.676$

$v_(1) \approx 14.607\,(m)/(s)$

The initial speed of the rock is approximately 14.607 meters per second.

b) We use (Eq. 1) once again and if we know that $g =9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $v_(1) \approx 14.607\,(m)/(s)$ and $v_(2) = 0\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-y_(2)) = (1)/(2)\cdot \left[\left(0\,(m)/(s) \right)^(2)-\left(14.607\,(m)/(s) \right)^(2)\right]$

$313.824-9.807\cdot y_(2) = -106.682$

$9.807\cdot y_(2) = 420.506$

$y_(2) = 42.878\,m$

The greatest height of the rock from the base of the cliff is 42.878 meters.

The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.

Answers

Answer:

σ = 0.8 N/m

Explanation:

Given that

L = 12 cm

We know that 1 m = 100 cm

L = 0.12 m

The force ,F= 0.096 N

Lets take surface tension = σ

We know that surface tension is given as

$\sigma =(F)/(L)\n\sigma =(F)/(L)\nNow\ by\ putting\ the\ values\n\sigma =(0.096)/(0.12)\ N/m\n\sigma=0.8\ N/m$

Therefore the surface tension σ will be 0.8 N/m .

σ = 0.8 N/m

Final answer:

The surface tension of the liquid in air is 0.8 N/m.

Explanation:

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

Learn more about Surface tension here:

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Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the UK where it is often very cloudy.

Answers

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

Mostly at night would they have been seen.
Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.

Answers

Answer:

42.67N

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

Testosterone is an example of what kind of biomolecule?

Answers

Answer:

Among the four biomolecules: carbohydrates, lipids, nucleic acids, and proteins it falls on the category of protein.

Explanation:

Testosterone, also known as 17-beta-hydroxy-4-androstene-3-one, is an androgen steroid hormone. It is largely released by the testes in males and the ovaries in females, although it is also secreted in minor amounts by the adrenal glands.

How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad/ s 2?