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Snezhnost [94]
2 years ago
7

Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This

coating leaves the threads, which are used to capture prey, hygroscopic - that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a 5-mm length of thread between two electrical contacts. The researchers stretched the thread in 1-mm increments to more than twice its original length, and then allowed it to return to its original length, again in 1-mm increments. Some of the resistance measurements are shown.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area A of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area of the coating compare when the thread is 13 long versus the starting length of 5 ? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.A13mm is about 1/10 A5mm.A13mm is about 1/4 A5mm. === correct answer... I figured it out. R = pL/A. L is 2.5 times. Therefore, A must be 1/4 times.A13mm is about 2/5 A5mm.A13mm is the same as A5mm.

Physics
1 answer:
Elodia [21]2 years ago
6 0

Answer:

A13 mm is about 1/4 A5 mm

Explanation:

Find the attachment

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Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

4 0
2 years ago
A pipe contains water at 500,000 Pa above atmospheric pressure. If you patch a 4.00 mm diameter hole in the pipe with a piece of
IRISSAK [1]

Answer: 6.284N

Explanation:

Pressure is the ratio of force exerted to cross sectional area of the material.

Pressure = Force/Area

Pressure = 500,000Pa

Area = Πd²/4 where d is the diameter of the hole.

If d = 4mm = 0.004m

Area = Π×0.004²/4

Area = 1.26×10^-5m²

Force = Pressure×Area

Force = 500,000× 1.26×10^-5

F = 6.284N

The gum must be able to withstand 6.284N force

7 0
2 years ago
What is the magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away?
sweet [91]
The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

F = kq1q2/r^2

where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters

Using direct substitution, the force F is determined to be 1920 Newtons.</span>
7 0
2 years ago
A physics professor that doesn’t get easily embarrassed stands at the center of a frictionless turntable with arms outstretched
mr Goodwill [35]
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.

L must stay the same before and after the professor brings the dumbbells closer to himself.

His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2

His final moment of inertia is 2.2kg•m^2.

Calculate the initial angular velocity:
L = 3.0π

Final angular velocity:
L = 2.2w

Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:

3.0π = 2.2w
w = 1.4π rad/s

The rotational energy is given by:
KE = 0.5Iw^2

Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J

Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J

There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
3 0
2 years ago
Read 2 more answers
Which statements describe the applications of nuclear medicine scans? Check all that apply.
coldgirl [10]

Answer:

- asses disease progression and tissue function

- utilize a biologically active molecule

- utilize a radionuclide tracer

Explanation:

4 0
2 years ago
Read 2 more answers
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