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mixas84 [53]
2 years ago
7

What is the approximate wavelength of a light whose second-order dark band forms a diffraction angle of 15.0° when it passes thr

ough a diffraction grating that has 250.0 lines per mm? 26 nm 32 nm 414 nm 518 nm
Physics
2 answers:
anzhelika [568]2 years ago
7 0
414nm just took the test
devlian [24]2 years ago
4 0
414 is the right answer. I just took the test and got an 100%
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A meteorologist tracks the movement of a thunderstorm with Doppler radar. At 8:00pm the storm was 55 mi northeast of her station
Rus_ich [418]
At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3
4 0
3 years ago
The conventional relatively small unit for work(ignoring time) such as raising one pound one foot is the foot-pound(ft. lb.). Si
Lelechka [254]

Answer:

1 joule = 0.737 foot-pound

Joule is the unit of work.

1 J = 1 N·m

In SI units

1 J = 1 kg· m/s²

0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.

6 0
2 years ago
Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other
denpristay [2]

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

3 0
3 years ago
Please help me with this question​
Ksju [112]

Answer:

i can not read that sorry

4 0
2 years ago
A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
2 years ago
Read 2 more answers
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