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melomori [17]
3 years ago
10

Newton's Laws question, please help me:

Physics
1 answer:
zaharov [31]3 years ago
6 0
I believe that would be the second law of motion
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Distinguish between nonsampling error and sampling error. Choose the correct answer below.
kiruha [24]

Answer:

The correct answer is D.

Non-sampling error is the error that results from​ under-coverage, non-response​ bias, response​ bias, or​ data-entry errors. Sampling error is the error that results because a sample is being used to estimate information about a population.

Explanation:

Sampling error is related to the variation between the true values of the sample and the population. If occurred, it is always random depending upon the sample chosen.

Non-sampling error can be random as well as non-random. Non-sampling error can occur irrespective of the sample chosen. It is related to the inappropriate analysis of the data.

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3 years ago
The path that thermal energy follows is from ___________ to ___________
Helga [31]
Warmer to colder objects
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3 years ago
Because of changes over time, the most accurate weather forecasts are: A) analog forecasts. B) long-term forecasts. C) seven-day
blondinia [14]
The answer is analog forecasts
8 0
2 years ago
Read 2 more answers
The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
Karolina [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

8 0
2 years ago
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The third-order bright fringe of 610-nm light is observed at an angle of 31° when the light falls on two narrow slits. How far a
Anna11 [10]

Answer:

The distance between the slits is 3.55 μm

Explanation:

Given that,

Order number = 3

Wave length = 610 nm

Angle = 31°

We need to calculate the distance between the slits

Using formula of distance of slit

d\sin\theta=n\lambda

d=\dfrac{n\lambda}{\sin\theta}

Where, n = order number

\lambda= wavelength

Put the value into the formula

d=\dfrac{3\times610\times10^{-9}}{\sin31}

d=3.55\times10^{-6}\ m

d= 3.55\ \mu m

Hence, The distance between the slits is 3.55 μm.

8 0
3 years ago
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