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Lady bird [3.3K]
3 years ago
6

- A thin film of oil * (n = 1.45) on a puddle of water, producing different colors. What is the minimum thickness of a place whe

re the oil creates constructive interference for light with a wavelength equal to 545 nm?

Physics
1 answer:
Darya [45]3 years ago
4 0

Answer:

There will be a phase change at the first interface and no phase change at the second interface:

If the film is 1/4 wavelength thick this restriction will hold

The wavelength of the light in oil is 545 nm / 1.45 = 376 nm

376 nm / 4 = 94 nm

"D" is correct

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horizontal force pushes block up a 20.0 incline with an initial speed 12.0 m//s. a) how high up the plane does slide before comi
Annette [7]

Answer:

Explanation:

We shall apply conservation of mechanical energy .

initial kinetic energy = 1/2 m v²

= .5 x m x 12 x 12

= 72 m

This energy will be spent to store potential energy . if h be the height attained

potential energy = mgh , h is vertical height attined by block

= mg l sin20  where l is length up the inclined plane  

for conservation of mechanical energy

initial kinetic energy  = potential energy

72 m = mg l sin20

l = 72 /  g  sin20

= 21.5 m

deceleration on inclined plane = g sin20

= 3.35 m /s²

v = u - at

t = v - u / a

= (12 - 0) / 3.35

= 3.58 s

it will take the same time to come back . total time taken to reach original point = 2 x 3.58

= 7.16 s

7 0
3 years ago
A car moves at a constant velocity of 30 m/s and has 3.6 × 105 J of kinetic energy. The driver applies the brakes and the car st
LekaFEV [45]

The force needed to the stop the car is -3.79 N.

Explanation:

The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.

So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be

K.E = \frac{1}{2} * m*v^{2}  \\\\m = \frac{2*KE}{v^{2} } = \frac{2*3.6*10^{5} }{30*30}=800 kg

Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,

Work done = Final kinetic energy - Initial kinetic energy

Work done = Force × Displacement

So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.

Force * 95 = 0-3.6*10^{5}\\ \\Force =\frac{-3.6*10^{5} }{95}=-3.79 N

Thus, the force needed to the stop the car is -3.79 N.

5 0
3 years ago
New attempt is in progress. Some of the new entries may impact the last attempt grading. Incorrect. The hammer throw is a track-
Morgarella [4.7K]

This question involves the concepts of centripetal force, range of projectile and projectile motion.

The magnitude of centripetal force is "2812.8 N".

First, we will find the velocity of the ball by using the formula of the range of the projectile.

R = \frac{v^2Sin2\theta}{g}

where,

R = range of projectile = 86.75 m

v = speed = ?

θ = launch angle = 47.9°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

86.75\ m = \frac{(v)^2Sin(2)(47.9^o)}{9.81\ m/s^2}\\\\v = \sqrt{\frac{(86.75\ m)(9.81\ m/s^2)}{Sin95.8^o}}

v = 29.25 m/s

Now, we will use the formula to find out the centripetal force:

F_c = \frac{mv^2}{r}

where,

F_c = Centripetal Force = ?

m = mass of the ball = 7.3 kg

v = speed = 29.25 m/s

r = radius = 2.22 m

Therefore,

F_c = \frac{(7.3\ kg)(29.25\ m/s)^2}{2.22\ m}

<u>Fc = 2812.8 N = 2.812 KN</u>

<u />

Learn more about centripetal force here:

brainly.com/question/11324711?referrer=searchResults

5 0
3 years ago
Which of the following is a true statement? A. Electromagnetic waves consist of only changing electric fields. B. Electromagneti
just olya [345]
C. Electromagnetic waves don't always need a medium to travel. Note that they do vary in wavelength and frequency however their speed is fixed. Also, EM waves are always transverse and they consist of vibrating electric and magnetic fields.
7 0
4 years ago
Read 2 more answers
A satellite orbits the Earth in the same direction it rotates in a circular orbit above the equator a distance of 250 km from th
UkoKoshka [18]

Answer:

Clock on the satellite is slower than the one present on the earth = 29.376 s

Given:

Distance of satellite from the surface, d = 250 km

Explanation:

Here, the satellite orbits the earth in circular motion, thus the necessary centripetal force is provided by the gravitation force and is given by:

\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}

where

v =  velocity of the satellite

R = radius of the earth = 6350 km = 6350000 m

G = gravitational constant = 6.674\times 10^{- 11} m^{3}/ks-s^{2}

M = mass of earth = 5.972\times 10^{24} kg

Therefore, the above eqn can be written as:

v = \sqrt{\frac{GM}{R}}

Now, for relativistic effects:

\frac{v}{c} = \sqrt{\frac{GM}{Rc^{2}}} = 26.41\times 10^{- 6}

Now,

r = R + 250

\frac{v_{surface}}{c} = {\frac{1}{c}\frac{2\pi R}{24} = 1.54\times 10^{-6}

Ratio of rate of satellite clock to surface clock:

\frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{surface}^{2}}{c^{2}}}} = 3.43\times 10^{-10}

Clock on the satellite is slower than the one present on the earth:

3.43\times 10^{-10}\times 24\times 3600 = 29.376 s

4 0
3 years ago
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