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Dima020 [189]
2 years ago
12

PLZ HURRY GOTTA HAVE IT DONE BEFORE MY MOM GETS HOME SHES 5 MINS AWAY!!!!!!!

Physics
1 answer:
Pie2 years ago
3 0

Answer:

The answer is B the products of photosynthesis are the reactants of cellular respiration

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Which describes Einstein’s second postulate about the special theory of relativity? The speed of light in a vacuum is constant,
crimeas [40]

Answer:A

Explanation:

5 0
2 years ago
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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

5 0
2 years ago
When did humans learn that the earth is not the center of the universe?
Vinvika [58]

Answer:

When did humans learn that the Earth is not the center of the universe?

Answer

1

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4 Answers

Asked in 3 Spaces





Science - Next Generation

Alexander Somm

, Consultant, Investor Relations at Novelpharm AG (2015-present)

Answered Oct 16

What, it isn’t?!

Sorry, I had to.

As far as I have read and understood, the Sumerians and later the Babylonians both had astronomical calendars that already differentiated planets and stars. Earth was not the center to them, the Sun likely was. That was around 2,200 - 1,600 BC.

After that, Greek philosopher Aristarchus of Samos (310 - 230 BC) was the first (recorded) to have believed the solar system was organized around the Sun, rather than the Earth. His heliocentric model was unpopular during Aristarchus’ lifetime, although it would inspire astronomers centuries later, such as Copernicus and Galileo.

Now, there are numerous archeological findings (cave paintings) and studies, that all suggest an understanding of complex astronomy in prehistoric times dating back as far as 40,000 years. This also explains how early, prehistoric migrants may have navigated the seas.

Explanation:

hope it helps

have a good day

4 0
2 years ago
Please help easy 8th grade question
Tamiku [17]

Answer:

whats the question?

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3 years ago
A wave with a period of 0.008 second has a frequency of
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The frequency of a wave is the reciprocal of its period.

A period of 0.008 sec means a frequency of

         1 / 0.008 sec  =  125 per sec .  (125 Hz)

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