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Dmitry_Shevchenko [17]
3 years ago
9

Stiind ca de la cea mai apropriata stea din afara Sistemului noatru Solar lumina ar ajunge in 4,3ani in ce unitati de masura con

siderati ca este potrivit sa se masoare distanta pana la aceasta stea?Ce valoare are aceasta distanta?
Physics
1 answer:
nevsk [136]3 years ago
3 0

Answer:

Petametres; 41 Pm

Explanation:

A measurement should use units that give a number between 0.1 and 1000.

The measure that astronomers use is the light year — the distance that light travels in one year.

In those units, the distance is 4.3 light years.

We can calculate the distance in SI units.

\text{Time} = \text{4.3 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} \times \dfrac{\text{24 h}}{\text{1 da}} \times \dfrac{\text{3600 s}}{\text{1h}} = 1.36 \times 10^{8} \text{ s}\\\\\text{Distance} = 1.36 \times 10^{8} \text{ s} \times \dfrac{2.998 \times 10^{8}\text{ m}}{\text{1 s}} = 4.1  \times 10^{16} \text{ m} = 41  \times 10^{15} \text{ m}

In SI, we should measure the distance in petametres (1 Pm = 10¹⁵ m)

Thus, the distance to the nearest star is 41 Pm.

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You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

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Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

8 0
3 years ago
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Which layer of the atmosphere is directly above the troposhpere? A.troposhpere B.stratosphere C.mesosphere D.exoshpere
vivado [14]
Its B. i hope i helped!!!
4 0
3 years ago
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The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque
Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

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4 0
2 years ago
Which of the following statements is TRUE for high-visibility clothing? A. High-visibility clothing helps to reduce insect probl
Vlad1618 [11]

Answer:

The answer for the above statement is:

C. High-visibility clothing is important to wear in areas with moving vehicles.

because in bright clothes you are easier to see, so people driving can see you.

Explanation:

3 0
3 years ago
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A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r
kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

\Phi = \frac{2\pi \delta}{\lambda}

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The horizontal distance between this two points would be given for

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\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}

\Phi= 1.096 rad \approx = 1.1 rad

Therefore the correct answer is C.

6 0
3 years ago
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