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omeli [17]
3 years ago
14

Which of the following is not found in living things? Which of the following is not found in living things?

Physics
2 answers:
ivann1987 [24]3 years ago
6 0
B would be the answer
inna [77]3 years ago
5 0

Answer:

PETROLEUM

Explanation:

THIS IS THE ANSWER BECAUSE ALL THE REST ARE FOUND IN LIVING THINGS

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We have to calculate how many square feet a school custodian has to wax and to choose the correct answer. A gymnasium floor is in the shape of the square ( 20 by 20 feet ). The area of the square is : A = L^2. For L = 20 ft, A = 20^2 = 20 * 20 = 400 ft^2. Answer: <span>D. 400 square feet.</span>
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What caused people to accept the geocentric model of the universe?
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I would say D, or B
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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in
valkas [14]

Answer:

Volume of the sample: approximately \rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}.

Average density of the sample: approximately \rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}.

Assumption:

  • \rm g = 9.81\; N \cdot kg^{-1}.
  • \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}.
  • Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to \rm 17.50 - 11.20 = 6.30\; N.

That's also equal to the weight (weight, m \cdot g) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with g.

\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg.

Assume that the density of water is \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}. To the volume of water displaced from its mass, divide mass with density \rho(\text{water}).

\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}.

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}.

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with g.

\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg.

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}.

3 0
3 years ago
A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu
marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

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3 years ago
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The answer to the question '<span>In a series of undisturbed rock layers where shale lies between sandstone below and limestone above, the __." would be limestone is youngest. Out of all the rock layers, the limestone is dated to be the youngest of all.</span>
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