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1 year ago

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An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o

ther words, what are the magnitudes of the components of the ship’s displacement vector in the north and the east direction, respectively?

1 answer:

Monica [59]1 year ago

6 0

I’m sorry if i took up a lot of space, hope this is a valid approximate answer

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A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through

Novay_Z [31]

Answer:

Angular spread, $\theta=1.472\times 10^{-7}\ rad$

Explanation:

It is given that,

Wavelength of the light, $\lambda=613\ nm=613\times 10^{-9}\ m$

Diameter of the telescope, D = 5.08 m

The minimum angular spread is given by :

$\theta=\dfrac{1.22\lambda}{D}$

$\theta=\dfrac{1.22\times 613\times 10^{-9}}{5.08}$

$\theta=1.472\times 10^{-7}\ rad$

So, the minimum angular spread of the beam is $1.472\times 10^{-7}\ radian$ . Hence, this is the required solution.

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A shield of material that reflects neutrons is placed around a radioactive sample. Which change in the nuclear reaction is most

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A chain reaction will be sustained in a sub-critical mass.

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1 year ago

Kinetic Energy Assignment: Lab Report

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Answer:

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2 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

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2 years ago