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VMariaS [17]
3 years ago
15

The magnetic fields in all planets are generated by the dynamo effect, caused by rapidly rotating and conducting material flowin

g in the interiors of the planets. We have just seen that all four of the jovian planets have magnetic fields. Earth also has a planetary magnetic field, but Venus does not, although it is very similar to Earth in mass. Which of the following would be a reason why Venus does not have a magnetic field?
Physics
1 answer:
exis [7]3 years ago
8 0

Answer:

Venus's very slow rotations rate does not allow a magnetic field to form

You might be interested in
A sports car accelerates from rest for 5 seconds reaching a velocity of 23.0 m/s.
denis-greek [22]

Answer:

<h2>4.6 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

<h3>a =  \frac{v - u}{t}  \\</h3>

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

Since the body is from rest u = 0

From the question we have

a =  \frac{23 - 0}{5}  =  \frac{23}{5}  \\

We have the final answer as

<h3>4.6 m/s²</h3>

Hope this helps you

4 0
2 years ago
Elements produce their spectrum when their electrons _____.
Aleonysh [2.5K]
The correct answer is the third one: move toward the ground state. Remember please that Elements produce their spectrum when their electrons move toward the ground state. Hope this is very useful
5 0
3 years ago
What is that distance traveled by an arrow going 78 m/s for 5 seconds
pshichka [43]

Explanation:

velocity=distance/time

distance= velocity×time

distance= 78×5=390m

8 0
2 years ago
Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti
Gwar [14]
After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it.  So only  (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter,  96.6%  of what entered it emerges from it, and
that's  96.6%  of  96.6%  of the original signal that entered the beginning
of the fiber.

==>  After 2 meters, the intensity has dwindled to  (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==>  After  'x'  meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
                                                     <em>(1,500) · (0.966)^x</em>
lumens of light remaining.
 
=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say  "15dB per 100 meters".

What does that mean ?    Break it down:  15dB in 100 meters is <u>0.15dB per meter</u>.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is  3.4% .  We applied
super high mathematics to that and calculated that  96.6% remains, or  0.966.

Look at this  ==>      10 log(0.966) =  <em><u>-0.15</u>  </em>  <==  loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in  'x'  meters of
fiber cable, doesn't have to mess with raising numbers to powers.  All he has to
do is say ...

--  0.15 dB loss per meter

--  'x' meters of cable

--  0.15x dB of loss.

If  'x' happens to be, say,  72 meters, then the loss is  (72) (0.15) = 10.8 dB .

and  10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083  =  <em>8.3%</em>  <== <u>That's</u> how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.
5 0
3 years ago
Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char
Dvinal [7]

(a) 2.98\cdot 10^{10} J

The change in energy of the transferred charge is given by:

\Delta U = q \Delta V

where

q is the charge transferred

\Delta V is the potential difference between the ground and the clouds

Here we have

q=31 C

\Delta V = 0.96\cdot 10^9 V

So the change in energy is

\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

K=\frac{1}{2}mv^2

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

K=2.98\cdot 10^{10} J

Therefore by re-arranging the equation, we find the final speed of the car:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s

5 0
3 years ago
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