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Alex17521 [72]
3 years ago
14

A parallel beam of light from a laser with a wavelength 450 nm, falls on a grating whose slits are 1.28 x 10^-4 cm apart. How fa

r (in cm) is the 1t order bright spot from the center of the pattern on a screen 3.4 m away?
Physics
1 answer:
Katena32 [7]3 years ago
7 0

Answer:

Y_1 = 1.195 m = 119.5 cm

Explanation:

For nth order bright fringe, the distance from the center of the pattern is given by :-

Y_n = \frac{n\lambda D}{d}.

Where,

n = order of nth  bright fringe. = 1

\lambda = wavelength of  given light  =  450 nm = 450*10-{9} m

D = distance between the screen and slits. =  3.4 m

and d = separation between the slits. = 1.28 *10^{-4} cm =1.28 *10^{-6} m

Y_1 = \frac{(1 *450 *10{-9} *3.4)}{(1.28*10{-6})}

Y_1 = 1.195 m = 119.5 cm

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A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.
DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy Q_{out}=1000\ Btu

Work done W_{cycle}=300\ Btu

We need to calculate the coefficient of performance

Using formula of  the coefficient of performance

COP=\dftrac{Q_{in}}{W_{cycle}}

We need to calculate the Q_{in}

W_{cycle}=Q_{out}-Q_{in}

Put the value into the formula

300=1000-Q_{in}

Q_{in}=300-1000

Q_{in}=700\ Btu

Now put the value of Q_{in} into the formula of COP

COP=\dfrac{700}{300}

COP=\dfrac{7}{3}=2.33

Hence, The coefficient of performance for the cycle is 2.33.

5 0
3 years ago
A soccer ball is kicked with a velocity of 8 m/s at an angle of 23°. what is the ball's acceleration in the vertical direction o
Aloiza [94]
Regardless of the speed of the ball or its angle, once it has left the kickers foot it's acceleration is always g downward. -9.81m/s^2
3 0
3 years ago
Read 2 more answers
A women does 80.0 joules of work when she slides a book 4.0 meters on the floor. How much force does she apply to the book
Kryger [21]
It would be 80/4 = 20 Newtons??? I think
5 0
3 years ago
A student throws a ball upward with a velocity of 35 m/s. What is the acceleration of the ball as it rises to the top of its arc
lora16 [44]

Answer:

The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

Explanation:

Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.

Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.

Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

7 0
3 years ago
A paper airplane is thrown to the north from the top of a cliff with a speed of 2 km/h. The paper airplane lands 15 minutes (0.2
Luden [163]
If you neglect air resistance, then you can solve for the horizontal distance using the formula:

dx=vixt

Where:
dx = horizontal distance
vix = initial horizontal velocity
t = time in flight

Now you can see that you have all the given you need. 
vix= 2km/h
t = 0.25 hours

We use hours because you need the units to be the same. So just plug that in your equation and you will have your answer:

dx = vixt
     =(2km/h)(0.25h)
     =0.5km

This means that the answer to your question is B.
5 0
3 years ago
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