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Agata [3.3K]
3 years ago
9

How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)

Physics
1 answer:
Svetach [21]3 years ago
4 0

To solve this problem, we will define speed as the amount of distance traveled per unit of time. We will clear the value of time and in parallel we will convert the Units to an international system to facilitate the calculation since we know the speed of light in a vacuum. The speed defined in terms of distance and time would be

v = \frac{x}{t} \rightarrow t = \frac{d}{v}

The speed of light in a vacuum is  3 * 10 ^ 8m / s and 1ft is equivalent to 0.3048m, so the estimated time would be

t = \frac{d}{v}

t = \frac{0.3048m}{3*10^8m/s}

t = 1.02*10^{-9}s

We know that 1 second is equivalent to 10 ^ 9ns, therefore

t =  1.02*10^{-9}s (\frac{10^9ns}{1s})

t = 1.02ns

Therefore the time is 1.02ns

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hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​
fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2      df = final position = 0 (on the ground)

                                           do =original position = 2 m

                                            vo = original <u>VERTICAL</u> velocity = 0

                                            a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t  - 1/2 ( 9.81)t^2

  to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

      10 m/s * .639 s =<u> 6.39 m </u>

5 0
1 year ago
Which best describes the forces identified by Newton’s third law of motion?
egoroff_w [7]
<span>equal and acting on different objects</span>
4 0
3 years ago
Read 2 more answers
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
3 years ago
The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able
vovangra [49]

Answer:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

q_e=-1.6\cdot 10^{-19}\ C

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C

Let's test the possible charges listed in the question:

-8.0 \times 10 ^{-19 }. We have just found it's a possible charge of a particle

-3.2 \times 10 ^{-19 }. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

-1.2 \times 10 ^{-19 } this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 } cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge -4.8 \times 10 ^{-19 }\ C is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

5 0
3 years ago
What are the forces like in a collision?
Anuta_ua [19.1K]

Answer:

The forces are a convergent

Explanation:

6 0
3 years ago
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