PHYSICS COLLEGE

Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?

Answers

Answer 1

Answer:

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, $q_1=-6.29* 10^(-6)\ C$

Second charged particle, $q_2=5.23* 10^(-6)\ C$

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

$F=k(q_1q_2)/(d^2)$

$F=9* 10^9* (-6.29* 10^(-6)* 5.23* 10^(-6))/((0.0359)^2)$

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

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If anyone can help with this, before 8 am eastern time i’ll give you 20$. Thank you

Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.

(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?

Answers

Answer:

a) $\lambda=2.95x10^(-6)m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=(c)/(\lambda)$

If we replace the last equation into the E formula we got:

$E=h(c)/(\lambda)$

And if we solve for $\lambda$ we got:

$\lambda =(hc)/(E)$

Using the value of the constant $h=4.136x10^(-15) eVs$ we have this:

$\lambda=(4.136x10^(15)eVs (3x10^8 (m)/(s)))/(0.42eV)=2.95x10^(-6)m$

$\lambda=2.95x10^(-6)m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^(-6)m$

A ball connected to a 1.1 m string and is swing in circular fashion. It’s tangential velocity is 15 m/s. What is its centripetal acceleration?

Answers

Answer:

ac = 204 [m/s²]

Explanation:

To solve this problem we must use the following equation that relates the tangential velocity to the radius of rotation.

ac = v²/r

where:

v = tangential velocity = 15 [m/s]

r = radius = 1.1 [m]

Now replacing we have:

ac = (15)²/1.1

ac = 204 [m/s²]

Which of the following sets of characteristics describe what we know about the outer planets? (2 points)Lowest daily average temperature, smaller in size and few or no moons.
Many moons, smaller in size and a ring system.
Rocky surface, closest to the sun and larger in size,
Gaseous composition, larger size and many moons.

Answers

Gaseous composition, larger size and many moons describe about the outer planets.

What are outer planets?

Jupiter, Saturn, Uranus, and Neptune are the four outer planets. They are all gas giants consisting primarily of hydrogen and helium. Their interiors are liquid and contain thick gaseous outer layers. Numerous moons and planetary rings consisting of dust and other particles are present on every one of the outer planets.

To know more about outer planets refer to :

brainly.com/question/18392639

#SPJ2

Answer: D

Explanation: Gaseous composition, larger size and many moons

the period of the satellite is exact 42.391 hours, the earth's mass is 5.98 kg and the radius of th earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles?

Answers

Answer:

As the mass is not written well, i will use the equation in terms of the gravitational acceleration:

The equation for the period of a satellite is:

$T = 2*pi*\sqrt{(r^3)/(g*R^2) }$

We want to find r, so isolating r we get:

$r = \sqrt[3]{x ((T)/(2*pi) )^2*g*R^2}$

Where:

T = period.

r = radius of the satellite.

R = radius of the planet.

g = gravitational acceleration of the planet.

pi = 3.14159...

g = 78999.64 mi/h^2 (value of a table)

T = 42.391 h.

R = 3958.8 miles

We can replace those values in the equation and get:

$r = \sqrt[3]{ ((42.391)/(2*3.14159) )^2*78999.64*(3958.8)^2} = 38,339.5 mi$

Now this value is measured from the center of the Earth, then the altitude of the satellite measured from the surface of the Earth will be:

H = r - R = 38,339.1mi - 3958.8mi = 34,380.3 mi

Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by magnetic forces. Michelle is a(n)