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Vesna [10]
3 years ago
10

The human eye can readily detect wavelengths from about 400 nm to 700 nm. part a if white light illuminates a diffraction gratin

g having 750 lines/mm , over what range of angles does the visible m = 1 spectrum extend?
Physics
1 answer:
Lera25 [3.4K]3 years ago
3 0

We are given that the wavelength ʎ is from 400 nm to 700 nm. The formula for this is:

d sin a =m * ʎ

where,

d = slit separation = 1 mm / 750 lines = 1/750

a = angle

m = 1

ʎ = 400 nm to 700 nm = 0.0004 mm to 0.0007 mm

 

Rewriting the formula in terms of angle a:

a = sin^-1 (m ʎ / d)

 

when ʎ = 0.0004 mm

a = sin^-1 (0.0004 / (1/750))

a = 17.46°

 

when ʎ = 0.0007 mm

a = sin^-1 (0.0007 / (1/750))

a = 31.67°

 

Hence the range of angles is from 17.46° to 31.67<span>°.</span>

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During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa
maw [93]

Answer:

Explanation:

Given

Initial speed u=60\ mi/hr\approx 88\ ft/s

distance traveled before coming to rest d_1=120\ ft

using equation of motion

v^2-u^2=2as

where v=final velocity

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0-(88)^2=2\times a\times 120---1

for u_2=80\ mi/hr\approx 117.33\ ft/s

using same relation we get

0-(117.33)^2=2\times a\times (d_2)----2

divide 1 and 2 we get

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A piece of blue paper appears blue because the paper
vfiekz [6]

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An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec
aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

x = V*t = -8\frac{m}{s} *3s = -24m

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

v^2 = v_0^2 + 2a*d

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s

Then, the time of the second phase will be:

t_2 = 9s - t_1 = 9s - 1.143s = 7.857s

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

x = \frac{1}{2}a*t^2 + v_0*t + x_0

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m

So, the total distance covered by this object in meters will be the sum of all the distances we found:

x_total = 24m + 4.57m + 216.07m = 244.64m

8 0
3 years ago
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