The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J
For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:

where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:

where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:

- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:

- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

- Replacing by the givens in (5), we can solve for μsmín, as follows:

Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
<span>The atom becomes positively charged.
When you add electrons to a neutral atom, it is no longer a neutral atom, it has a negative change and is an anion. When you take away electrons from a neutral atom, it is no longer a neutral atom- it becomes a positive atom, and is a cation.</span>