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NNADVOKAT [17]
2 years ago
6

A nanometer is a unit of mass, whereas a nanosecond is a unit of time. Question 1 options: True False

Physics
1 answer:
Alenkasestr [34]2 years ago
6 0

Answer:

True.

Explanation:

A nanometer is a unit of mass, whereas a nanosecond is a unit of time. To convert 1.3 hours to minute, you would multiply by 1 h / 60 min. Kilometer is a unit of length, whereas kilogram is a unit of mass. True.

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A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0 nm and 589.6 nm. The
DIA [1.3K]

The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

<u><em>L = 57.88 mm</em></u>

<u><em /></u>

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

Read more at; brainly.com/question/17161594

8 0
2 years ago
8. Noticing how much food is on your lunch<br> tray is a quantitative observation because
zhenek [66]

Answer:

you are making an observation that uses numbers.

Explanation:

5 0
2 years ago
Is how high or low sound is
JulsSmile [24]

Answer:

How high the sound or how low the sound is depending on the pitch of the sound which in this case the frequencies of the sound. The higher of the frequencies , the higher of the pitch but it has the shortest length of wave (λ). That's why AM radio have a longer range but bad audio quality than FM radio that have better audio quality with shorter range.

5 0
3 years ago
A 70kg man is moving with a speed of 3.2 m/s on a rough surface, his speed was decreasing uniformly until he reaches to a stop b
VMariaS [17]

Answer:

224 N

Explanation:

6 0
3 years ago
Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage
agasfer [191]

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

Therefore, the velocity at which the exhaust gases exit the engines is

v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

6 0
2 years ago
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