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Vinil7 [7]
2 years ago
7

What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angl

e of 60.0º
Physics
1 answer:
Nataly [62]2 years ago
8 0
The condition for maximum intensity is the same as that for a double slit. However, angular separation of the maxima is generally much greater because the slit spacing is so small for a diffraction grating. For a diffraction grating with lines/mm = lines/inch, the slit separation is d = micrometers = x10^ m.
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A first-order reaction has a rate constant of 0.241/min. if the initial concentration of a is 0.859 m, what is the concentration
nordsb [41]

Answer: 0.077 M

Explanation:

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant = 0.241minute^{-1}

t = time taken for decay process = 10 minutes

a = initial amount of the reactant= 0.859 M

a - x = amount left after decay process =?

Putting values in above equation, we get:

0.241 minutes^{-1}=\frac{2.303}{10.0}\log\frac{0.859}{a-x}

(a-x)=0.077M

Thus the concentration of a after 10.0 minutes is 0.077 M.


5 0
3 years ago
Read 2 more answers
An Amtrak going 250m/s comes to a stop in 12s. What is the<br> acceleration?
astraxan [27]

Answer:

a=\frac{v-u}{t}  \\a = \frac{0-250}{12} = -20.83 m/s

Explanation:

you mean deceleration right ? because the acceleration is 250m/s

7 0
3 years ago
A crane lifts a 400 N crate full of baked beans 15 m. How much work was done by the crane?
damaskus [11]

Answer:

6000 J

Explanation:

work = F x S

work = 400 x 15

work = 6000 J

3 0
3 years ago
. Consider a fully extended arm that is rotating about the shoulder such that with a shoulder-to-hand length of 30 cm. If the ar
Rainbow [258]

Answer:

13.309 m/s²

Explanation:

Length from shoulder to hand, l = 30 cm = 0.3 m

initial velocity, u = 1 m/s

final velocity, v = 2.5 m/s

time, t = 3 s

Let the tangential acceleration is a.

by using first equation of motion

v = u + at

2.5 = 1 + 3 a

a = 0.5 m/s²

Let the centripetal acceleration is a'.

a' = v'²/l

a' = 2 x 2 / 0.3

a' = 13.3 m/s²

The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by

A=\sqrt{a^{2}+a'^{2}}

A=\sqrt{0.5^{2}+13.3^{2}}

A = 13.309 m/s²

3 0
3 years ago
Which of the following is not a part of the appendicular skeleton
Bingel [31]

Answer:

Hyoid

Explanation:

The hyoid is located in the neck area, not the limbs.

7 0
3 years ago
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