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nlexa [21]
3 years ago
10

Look at the image if you need me to add the answers so you can see them better i will

Physics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

0.75 meters per second; north

Explanation:

speed (velocity)= distance/time

15 meters= distance

20 seconds= time

15/20= 0.75 meters per second; north

Hope this helps!

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A conducting spherical shell with inner radius a and outer radius b has a positive point charge +Q at the center in the empty re
Andreas93 [3]

Answer:

a) q_inner = -Q

, q_outer = -2Q

b)    E₁ = k Q / r²        r<a

       E₂ = 0               a<r<b

       E₃ = - k 2Q/r²     r>b

d)   the charge continues inside the spherical shell, the results do not change

Explanation:

a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign  

q_inner = -Q  

the outer shell of the shell the load is  

q_outer = -3 Q + Q  

q_outer = -2Q

b) To find the electric field again, use Gauss's law,  

We define as a Gaussian surface a sphere  

Ф = E. dA = q_{int}/ε₀

in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product  

E A = q_{int}/ε₀

the area of ​​a sphere is  

A = 4 π r²  

E = 1 / 4πε₀  Q/ r²  

k = 1 / 4πε₀  

let's apply this expression to the different radii

i) r <a  

in this case the load inside is the point load  

q_{int}= + Q  

E₁ = k Q / r²

ii) the field inside the shell  

a <r <b  

As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.  

E₂ = 0

iii)  r>b

   q_{int}  = Q- 3Q = -2Q

    E₃ = k (-2Q/r²)

     E₃ = - k 2Q/r²

c) see  attached

d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface

6 0
3 years ago
132 projects in science fair. if 8 projects fit in a row. how many full rows
Umnica [9.8K]

The real answer would be 16.5 but since you want to have a full number it would be 16

3 0
3 years ago
Read 2 more answers
An example of a land feature formed by river erosion is?
KatRina [158]
<span>There are various examples of land features formed by river erosion. One of them is Delta. A delta is formed when sediments are deposited in a place where a river flow into an ocean or lake and build up a land form. Thus, a river delta is a land form that occurs as a result of deposition of sediments carried by a river as the flow leaves its mouth and enter a slow moving or stagnant water.</span>
7 0
3 years ago
The incoming infrared radiation from the Sun as it travels towards Earth is _____. mostly reflected back out into space mostly a
s2008m [1.1K]

Answer:

Mostly absorbed and transmitted toward Earth's surface

Explanation:

Infrared radiation from the sun reaches the Earth's surface and warms the Earth's surface. It emits some infrared radiation, some absorbed by gases such as carbon dioxide in the atmosphere. This is known as the greenhouse effect. and If it don't, radiation from the earth is powerful enough to burn the skin.

so correct answer is Mostly absorbed and transmitted toward Earth's surface

7 0
3 years ago
Read 2 more answers
A satellite in orbit around the earth has a period of one hour. An identical satellite is placed in an orbit having a radius tha
Advocard [28]

Answer:

27h

Explanation:

We can answer this problem with Kepler's third law:

\frac{T^2}{a^3} =constant

where T is the period and a is the semi-major axis of an ellipse, but because the satellite's orbit is around the earth it must be a circular orbit, so that a becomes the radius r of a circle.

\frac{T^2}{r^3} =constant

what this tells us is that the relationship between the first period with radius r must be equal to the second period when the radius is now 9 times the original (the second radius is 9r because the orbit is nine times larger):

We will have the following:

\frac{T_{1}^2}{r^3} =\frac{T_{2}^2}{(9r)^3}

T_{1} is the original period: 1 hour

T_{2} is the period of the second satellite.

and (9r)^3=729r^3

thus:

\frac{(1h)^2}{r^3} =\frac{T_{2}^2}{729r^3}

Clearing for T_{2}

\frac{(1h)^2}{r^3}(729r^3) ={T_{2}^2}

729=T_{2}^2

\sqrt{729}=T_{2}

27=T_{2}

The period of the second satellite is 27 hours

6 0
3 years ago
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