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blsea [12.9K]
3 years ago
5

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

0 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th dark fringe
Physics
1 answer:
cluponka [151]3 years ago
8 0

Answer:

The thickness is  \Delta y =  2.4 *10^{-6} \  m

Explanation:

From the question we are told that

   The wavelength is  \lambda  = 480 \ nm = 480*10^{-9} \  m

    The first order of the dark  fringe is  m_1 =  16

     The second order of dark fringe considered is  m_2 = 6

Generally the condition for destructive interference is mathematically represented as

        y = \frac{m \lambda}{2}

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      y_1 - y_2 = \Delta y =  \frac{m_1 \lambda}{2} -  \frac{m_2 \lambda}{2}

=>  y_1 - y_2 = \Delta y =  \frac{16 *480*10^{-9}}{2} -  \frac{6 *480*10^{-9}}{2}

=>  y_1 - y_2 = \Delta y =  5 (480*10^{-9})

=>  \Delta y =  2.4 *10^{-6} \  m

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ValentinkaMS [17]

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

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3 years ago
How many suns do you think are in our galaxy? Explain your answer PLS HELP ASP BEST GETS BRAINLIEST​
dexar [7]

Answer: There is only one Sun in the galaxy … that is the thing that rises in the morning and sets at night. However, there is a use of “sun” to signify any old star … nobody knows exactly there might be trillions out there

Explanation:

3 0
3 years ago
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

5 0
3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
6. 498.82 mg comverted to kg​
Ivanshal [37]

Answer:

0.00049882 kg

........................

6 0
3 years ago
Read 2 more answers
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