How long does it take for mars to rotate on its axis
2 answers:
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Answer:
<em> -18896.49 V/m</em>
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Explanation:
Distance between the two plates = 10 cm = 10 x m = 0.1 m
Also, one of the plates is taken as<em> zero volt.</em>
a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V
b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V
<em>Potential field strength = -dV/dx</em>
where dV is voltage difference between these points,
dx is the difference in distance between these points
For the first case above,
potential field strength = -393/0.0705 = -5574.46 V/m
For the second case ,
potential field strength = -393/0.0295 = -13322.03 V/m
Magnitude of the field strength across the plates will be
-5574.46 + (-13322.03) = -5574.46 + 13322.03 =<em> -18896.49 V/m</em>
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
Box with mass 2M
Box with mass 3M
On the third equation, acceleration can be modelled in terms of F'':
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
Afterwards, F' as function of the external force can be obtained by direct substitution:
The net forces of each block are now calculated:
Box with mass M
Box with mass 2M
Box with mass 3M
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.