Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of

Answer:

The plane's speed in relation to the ground is 300.79 km/h.

Explanation:

Provided details include:

Wind speed = 75.0 km/hr

Plane's airspeed = 310 km/hr

Next, we must find the ground speed of the plane

Calculating the angle

Using the angle formula

Where v' represents the wind speed

v represents the plane's speed

We will substitute the values into the formula

Now, we must find the resultant speed

Using the resultant speed formula

Insert the values into the formula

Consequently, the plane's speed in relation to the ground equals 300.79 km/h.

The masses of particle A, B, and C are given, with all three particles aligned linearly. The distances between them are noted. The gravitational forces are attractive, compounding when acting in the same direction. The effects on each particle are formulated based on their distances.

The amount of work performed by a system at consistent pressure is defined by the following equation:

where
p represents pressure
as the final volume
as the initial volume

Plugging the values given in this case into the formula gives us


Considering that , the result for the work done becomes

The tension exerted in the cable amounts to T = 16653.32 N. Parameters: Cross section area A = 1.3 m² Drag coefficient CD = 1.2 Velocity V = 4.3 m/s The angle formed by the cable with the horizontal is 30 degrees. Density defined as follows: The drag force FD is determined by the equation: FD = (1/2) * ρ * V² * A * CD Calculating the drag force yields 14422.2 N acting opposite to motion. Given the cable's angle of 30 degrees with horizontal, the horizontal component contributes to the drag force calculation: T * cos(30) = F_D Thus, T = 16653.32 N.

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k