Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Answer:
The correct option is;
B) No, the Navy vessel is slower
Explanation:
The speed of some torpedoes can be as high as 370 km/h. The average speed of a fast Navy vessel is approximately 110 km/h
Therefore, the torpedoes travel approximately 3 times as fast as the (slower) Navy vessel, such that the torpedo covers three times the distance of the Navy vessel in the same time and therefore, if the Navy vessel and the torpedo continue in a straight line (in the same direction) due north the vessel can not outrun the torpedo
Therefore, no the Navy vessel travels slower than a torpedo.
The answer is 3. Photosynthesis removes carbon dioxide while respiration puts back carbon dioxide
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
