Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
The maximum number of bright spot is
From the question we are told that
The slit distance is
The wavelength is
Generally the condition for interference is
Where n is the number of fringe(bright spots) for the number of bright spots to be maximum
given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as
The 1 here represented the central bright spot
HELP ME PLS!!!!Find the location of beryllium (Be) on the periodic table. What type of ion will beryllium form? A. An ion with a -2 charge B. An ion with a +6 charge C. An ion with a +2 charge D. An ion with a -6 charge
The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.
What are cations and anions?
In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.
This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.
Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.
The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.
Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.
Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.
A celestial body moving in an ellipical orbit around a star
A celestial body moving in an elliptical orbit around a star is a planet.
Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.
A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?
The change in momentum is
From the question we are told that
The mass of the probe is
The location of the prob at time t = 22.9 s is
The momentum at time t = 22.9 s is
The net force on the probe is
Generally the change in momentum is mathematically represented as
The initial time is 22.6 s
The final time is 22.9 s
A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.
The answers to the questions are;
(a) The velocity of the truck right after the collision is 20.884 m/s
(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J
(c) The change in mechanical energy is due to energy consumed by the collision process.
(a) From the principle of conservation of linear momentum, we have
m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄
m₁ = Mass of the car = 1225.0 kg
m₂ = Mass of the truck = 9700.0 kg
v₁ = Initial velocity of the car = 25.000 m/s
v₂ = Initial velocity of the truck = 20.000 m/s
v₃ = Final velocity of the car right after collision = 18.000 m/s
v₄ = Final velocity of the truck right after collision
1225.0 kg × 25.000 m/s + 9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s + 9700.0 kg × v₄
That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄
Making v₄ the subject of the formula yields
v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s
The velocity of the truck right after the collision to five significant figures = 20.884 m/s
(b) The change in mechanical energy of the car truck system in the collision can be found by
The change in kinetic energy of the car truck system
Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy