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Lunna [17]
3 years ago
12

A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The

fifth-order maximum of the unknown wavelength exactly overlaps the third-order maximum of the red light.
What is the unknown wavelength? (in nm)
Physics
1 answer:
marusya05 [52]3 years ago
4 0

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

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The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles
egoroff_w [7]

60 mph east...........

6 0
3 years ago
Read 2 more answers
A car moves along a curved road of diameter 2 km. If the maximum velocity for safe driving on this path is 30 m/s, at what angle
Leto [7]
The maximum velocity in a banked road, ignoring friction, is given by;

v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.

Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°

Therefore, the road has been banked at 5.24°.
4 0
3 years ago
Electromagnetic waves are ........... by shiny surfaces.
sleet_krkn [62]

Answer:

Electromagnetic waves are reflected

Explanation:

Reflection of light (and other forms of electromagnetic radiation) occurs when the waves encounter a surface or other boundary that does not absorb the energy of the radiation and bounces the waves away from the surface. ... This concept is often termed the Law of Reflection.

4 0
3 years ago
An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?
Luden [163]

Answer:

a=2.8\ m/s^2

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2

So, the acceleration of the object is 2.8\ m/s^2.

7 0
3 years ago
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