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3 years ago

Which best describes accuracy?

Physics

2 answers:

zysi [14]3 years ago

4 0

Answer:

the agreement between a measured value and an accepted value

Explanation:

Vitek1552 [10]3 years ago

3 0

Answer: A.) the smallness of the graduations on a measuring tool

Explanation:

Accuracy is defined as the closeness of the measured value to the actual value. Therefore, out of the options, this is best described by A. the smallness of the graduations on a measuring tool. This is because smaller graduations mean that the value can be measured to a smaller level, making it more accurate. PLEASE GIVE ME BRAINIEST

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The world's largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds. What is the velocity fo

Arturiano [62]

Answer:

The velocity of the blades is 88.185 m/s.

Explanation:

Given;

length of the blade, r = 80 m

angular speed, ω = 1 rev per 5.7 seconds

The velocity of the blades is calculated by applying the following circular motion equation that relates linear velocity (V) and angular speed (ω);

$V = \omega r\\\\V = (\frac{1 \ rev}{5.7 \ s} \times \frac{2 \pi \ rad}{ 1 \ rev} )(80 \ m)\\\\V = 88.185 \ m/s$

Therefore, the velocity of the blades is 88.185 m/s.

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3 years ago

How does gravity affect maglev trains?

yuradex [85]

Earth pulls it downward to the gravitational force

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3 years ago

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Snowcat [4.5K]

Answer:

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Explanation:

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3 years ago

The group of test subject that are NOT given the experimental treatment is called what?

-BARSIC- [3]

Answer:

The group that remains unaltered is called the control group.

5 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago