6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough
Answer:
It travels 1414 feets.
Explanation:
Let's take the length the bullet travels <em>l </em>as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height <em>h</em> it reached, we can calculate <em>l</em> using the <em>sin(α)</em> function:

Replacing:

Solving and roundin to the nearest foot:

Answer:
The value of radiation pressure is
Pa
Explanation:
Given:
Intensity

Area of piece

From the formula of radiation pressure in terms of intensity,

Where
radiation pressure,
speed of light
We know value of speed of light,

Put all values in above equation,

Pa
Therefore, the value of radiation pressure is
Pa
Answer:
(a) 5.04 eV (B) 248.14 nm (c) 
Explanation:
We have given Wavelength of the light \lambda = 240 nm
According to plank's rule ,energy of light


Maximum KE of emitted electron i= 0.17 eV
Part( A) Using Einstien's equation
, here
is work function.
= 5.21 eV-0.17 eV = 5.04 eV
Part( B) We have to find cutoff wavelength



Part (C) In this part we have to find the cutoff frequency

An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.