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navik [9.2K]
3 years ago
10

a point charge q=-0.35 nc is fixed at the origin. where must an electron be placed in order for the electric force acting on it

to be exactly opposite its weight?
Physics
1 answer:
Tema [17]3 years ago
8 0

Answer:

238 km above q

Explanation:

Assuming no other external forces present, the electron must be in equilibrium, between gravity (always downward) and the electrostatic force (given by Coulomb's Law) that must be upward.

As the force on the electron due to the negative charge in the origin is repulsive, the electron must be located above the charge q.

In order to find the distance to the origin, we know that the weight and the electrostatic force are equal each other:

Fe = W\\\frac{k*q*e}{d^{2}} = m*g\\d =\sqrt{\frac{k*q*e}{m*g} }  = \sqrt{\frac{9e9 N*m2/C2*0.35e-9C*1.6e-19C}{9.1e-31kg*9.8m/s2} } =237,697 m = 238 km

⇒ d= 238 Km above q.

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What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?
statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000  x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0
3 years ago
Estimate the total number of bacteria and other prokaryotes in the biosphere of the earth. (assume the bacteria are found to a d
Fofino [41]
Since the Earth is almost spherical in shape, we are actually to find first the volume of the spherical segment at a depth of 1,000 m. The radius of the Earth is 6,371,000 meters. The volume of a spherical segment is:

V = 1/3*πh²(3r - h)
Substituting the values and making sure the units is in mm,
V = 1/3*π(1000 m * 1000 mm/1 m)²[3(6,371,000 m * 1000 mm/1 m) - (1000 m * 1000 mm/1 m)]
V = 2×10²² mm³

Thus, the total amount of bacteria is:

2×10²² mm³ * 100 bacteria/1 mm³ = 2×10²⁴ bacteria
7 0
4 years ago
The electric current leaves the battery through the --- complete this blank
aev [14]
The bulb

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5 0
4 years ago
Why is an object's density expressed as a relationship between two units?
vfiekz [6]
Probably for kind of the same reason that speed is expressed as a
relationship between two units.  You know, like miles per hour .

I guess the only reason is because no single unit has been invented
to describe density.

The rate of doing work or using energy would always be expressed
as a relationship between two units ... we would say that the rate of
work is "(so many) joules per second".  But the "watt" was invented,
so we can say "(so many) watts" instead.

So I guess you're right.  Density could be simpler to describe
if we only had a unit for it.  Then we wouldn't have to say "(so many)
grams per cubic centimeter".  We would just say "(so many) (new unit)".

Let's try it out:
 
"Uhhh, pardon me Professor . . . I've been working late in the lab,
and I believe I've identified a new substance, hitherto unknown to
the scientific community, and totally unexpected.  In its pure form,
the substance appears to be pink, it smells like butterscotch, and
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I like it !
5 0
4 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
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