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disa [49]
2 years ago
15

How do I solve this?​

Physics
1 answer:
prohojiy [21]2 years ago
3 0

unlock your inner brain

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Which expression of distance uses SI units? A. 30 miles B. 16 kilograms C. 24 feet D. 500 meters
Stella [2.4K]
Hi!

SI units are physical measurements which will be in the form of kilograms, second, kelvin, metres, etc.

Since kilograms measure the weight of an object, it is out. Miles and feet are not SI units, so they are also out. This only leaves one answer left!

Hopefully, this helps! =)
7 0
3 years ago
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Communicating results allows scientists to learn from each other's results.
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True, scientists often talk to each other to figure out if their results were similar and what they could have done better.

Although, talking to other scientists does have risks, other scientists could copy your work and further better it.

So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol

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3 years ago
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In a stadium, fans stand up and sit down to produce a wave across the stadium. This type of wave where the material travels perp
puteri [66]
Correct option B

Transverse waves are those waves whose particles vibrate perpendicular to the direction of wave.

Hope This Helps You!
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3 years ago
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A bird flies 3.7 meters in 46 seconds, what is its speed?
victus00 [196]

Answer:

Speed is 0.08 m/s.

Explanation:

Given the distance that the bird flies = 3.7 meters

The time is taken by the bird to fly the 3.7 meters = 46 seconds  

We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.  

Below is the formula to find the speed.

Speed = Distance / Time

Now insert the given value in the formula.

Speed = 3.7 / 46 = 0.08 m/s

8 0
3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
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