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Mekhanik [1.2K]
3 years ago
11

Which of the following is NOT a unit of measurement for speed?

Physics
2 answers:
Art [367]3 years ago
6 0
I’m not sure what to go off of sorry
Dmitriy789 [7]3 years ago
3 0

I don't see any units of measurement or speed I don't see anything

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The Kelvin scale is also know as the absolute zero<br> TRUE OR FALSE
Bond [772]

The Kelvin scale is also called absolute zero scale

Explanation:

true

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How much does a bowling ball weigh if it has a mass of 6 kg on Earth?
larisa [96]

Answer: pretty sure 14 pounds

Explanation:

7 0
3 years ago
Refer to the following diagram to answer this question
svetoff [14.1K]
In this case, the resistance for the diagram shown in the figure is located at point D. The symbol that best represents a resistance in an RLC diagram is one that is similar to that of point D.
 Also, remember that by definition V = I * R.
 The answer is D.
7 0
4 years ago
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When it orbited the moon, the apollo 11 spacecraft's mass was 10400 kg, its period was 133 min, and its mean distance from the m
Anna11 [10]
Since the orbit is circular and you are given the period and average distance( in other words the radius of the orbit) we can calculate the orbital speed right away. Keep in mind that ship's mass is much smaller than the mass of the moon, this is really important, as we can simply say that moon is not affected by the ship. 
Angular frequency and orbital speed have the following relationship:
v=wr=\frac{2\pi}{T}\cdot r
v=\frac{2\pi}{133\cdot 60}1.99345\cdot 10^6=1569.5795\frac{m}{s}
Let's do the calculation with rest of the numbers just to make sure.
Because you have the stable circular orbit the gravitational force and centrifugal force are in balance:
G\frac{mM}{r^2}=m\frac{v^2}{r}
We solve this for v:
G\frac{M}{r}=v^2\\ v=\sqrt{G\frac{M}{r}}
You can use this formula for any celestial body as long as the ship's mass is much smaller than the mass of the body, and this only applies to circular orbits
When we plug in the numbers we get:
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As you can see the results are identical.
6 0
3 years ago
Two plates of area 30.0 cm^2 are separated by a distance of 0.0590 cm. If a charge separation of 0.0240 μC is placed on the two
topjm [15]

Answer:

V=533.33 V

Explanation:

 Given that

A= 30 cm²

d= 0.059 cm

Q= 0.0240 μ C

We know that capacitance given as

C=\dfrac{\varepsilon _oA}{d}

Now by putting the values

C=\dfrac{8.85\times 10^{-12}\times 30\times 10^{-4}}{0.059\times 10^{-2}}

C=4.5\times 10^{-11}\ F

Voltage difference given as

Q= V .C

V=Q/C

V=\dfrac{0.0240\times 10^{-6}} {4.5\times 10^{-11}}

V=533.33 V

3 0
3 years ago
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