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Brums [2.3K]
2 years ago
13

Which type of error is a false positive

Physics
1 answer:
maks197457 [2]2 years ago
5 0

Answer: Type I Error

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A quantity that is a measure of how the displacement of a body changes with time is:
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Explanation:

Speed

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What is the relationship between wavelength and amplitude? Best answer gets brainliest!
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wavelength refers to the length of a wave from one peak to the next.the amplitude of a wave Is measured from the peak to the trough

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10 kg mass sliding along a horizontal rough surface at speed 30 m/s and rests in 6 seconds. what is the coefficient of friction
Aleks [24]

The coefficient of friction is 0.5.

<h3>What is Coefficient of friction?</h3>

The ratio of friction force to normal force is known as the coefficient of friction (COF), which has no dimensions. Those materials are said to as lubricous if their COF is less than 0.1. Surface roughness and COF are dependent on the composition of the materials.

Initial velocity, u=30m/s

time taken = 6s

Final velocity, v=0

Acceleration of friction, a=μg=10μms⁻²

Apply first kinematic equation of motion,

v - u = at

0 - 30 = (-10μms⁻²) × 6

-30 = -60μ

μ = 1/2

μ = 0..5  

Hence, coefficient of friction is 0.5

to learn more about Coefficient of friction go to - brainly.com/question/14121363

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2 years ago
Explain one way the water cycle affects climate. Use complete sentences.
Reika [66]
When water vaporizes into the air, it becomes humid out.
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The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If
ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

y=3+0.8x-0.4x^2........(1)

The x component of constant velocity, v_x=5\ m/s

We need to find the resultant velocity at the point (2,3).

Let \dfrac{dx}{dt}=v_x and \dfrac{dy}{dt}=v_y

Differentiating equation (1) wrt t as,

\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}

v_y=0.8\times v_x-0.8x\times v_x

v_y=0.8v_x(1-x)

When x = 2 and v_x=5\ m/s

So,

v_y=0.8\times 5\times (1-2)

v_y=-4\ m/s

Resultant velocity, v=\sqrt{v_x^2+v_y^2}

v=\sqrt{5^2+(-4)^2}

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0
3 years ago
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