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horrorfan [7]
3 years ago
9

How significant is air resistance for projectile motion?

Physics
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

Air resistance have some significant role in projectile motion if the motion lasts for some time.

Explanation:

  • Air resistance or air drag seems to be important in daily actvities and games like baseball.
  • The trajectory of the projectile with or without air resistance or air drag is totally different.
  • When we neglect air drag, the only acting force is gravity against the motion so the maximum height and range are suppose Hmax and R.
  • Now, when we consider air drag, it is important to notice that there are two forces against the motion of the ball and along the direction of gravity. It seems that both maximum height and range are lesser Hmax'< Hmax and R'<R.
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Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p
Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation  that is

for red light

h ( c / λr ) = Ф +  kinetic energy

for green light

h ( c / λg ) = Ф +  1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф +  1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr  - 1/ λg)

put all value

Ф =( 6.63 ×10^{-34} (3 ×10^{8} )  / 0.50) × ( 1.50/ λr  - 1/ λg)

Ф =( 6.63 ×10^{-34} (3 ×10^{8} ) / 0.50 ) × ( 1.50/ 651×10^{-9}   - 1/ 521 ×10^{-9})

Ф = 1.5305  ×10^{-19} J  × ( 1ev / 1.6 ×10^{-19} J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0
3 years ago
Read 2 more answers
PLZ HELP NO SNEAKY LINK OR I WILL REPORT U OR NON ANSWERS
Lana71 [14]
Core
Home of atoms of hydrogen also the lightest element in the universe.
Radiative Zone
Outside the inner Core it radiates energy through the process of photon emission.
Convection Layer
Outer most Layer of the Core, it extends form a depth of 200,000 kilometres to the visible surface. Energy is created by Convection. This is where light is produced.
Photosphere
Surrounds the stars and is where light and heat radiate.
Chromosphere
Reddish gas layer outside of the photosphere I think it also works with the Corona.
Corona
Aura of Plasma that surrounds the Sun and other stars, it extends millions of kilometres and easily seen during a total eclipse.
8 0
3 years ago
Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy
mylen [45]

Answer:

x_total = (A + B) cos (wt + Ф)

we have the sum of the two waves in a phase movement

Explanation:

In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.

         Max

               x₁ = A cos (wt + Ф)

         Jimmy

              x₂ = B cos (wt + Ф)

         

total movement

             x_total = (A + B) cos (wt + Ф)

 Therefore we have the sum of the two waves in a phase movement

8 0
3 years ago
This company is run by kids right
GREYUIT [131]

Answer:

i d k about that but I know it`s a Polish thing

4 0
3 years ago
Read 2 more answers
The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-
WINSTONCH [101]

Answer: 1.212(10)^{-10} m

Explanation:

The de Broglie wavelength \lambda is given by the following formula:

\lambda=\frac{h}{p} (1)

Where:

h=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

p is the momentum of the atom, which is given by:

p=m_{e}v (2)

Where:

m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg is the mass of the electron

v=6(10)^{6}m/s is the velocity of the electron

This means equation (2) can be written as:

p=(9.11(10)^{-31}kg)(6(10)^{6}m/s) (3)

Substituting (3) in (1):

\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)} (4)

Now, we only have to find \lambda:

\lambda=1.2122(10)^{-10} m>>> This is the de Broglie wavelength of the electron

8 0
3 years ago
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