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artcher [175]
3 years ago
15

What wires allow electrons to flow the easiest?

Physics
1 answer:
docker41 [41]3 years ago
5 0
The shortest and straight ones
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How long did Skylab orbit the earth?
maria [59]

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for abt 36 seconds

Explanation: cuz thats how long it was

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Which "spheres" are interacting when water evaporates from plants
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The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.
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Which of the following best describes wind?
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The answer is D the rising of warm air pushing down cool air.

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These are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the oz
yarga [219]
Hello! The nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the ozone layer are the Chlorofluorocarbons (CFCs)

These are compounds developed and improved by Thomas Midgley in the late 1920s. They were used as refrigerants and aerosol propellants.

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4 0
4 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
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