Answer:
magnetic field will allow the electron to go through 2 x T k
Explanation:
Given data in question
velocity = 5.0 ×
electric filed =
To find out
what magnetic field will allow the electron to go through, undeflected
solution
we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions
so we can also say
F(net) = Fe + Fb i.e. = 0
q V B + q E = 0
q will be cancel out
+ 5e + 7i × B = 0
B = 2 x T k
Answer:
The correct answer is:
a) remain where it is released
Explanation:
The concept of density seeks to measure the weight of an object in relation to its size. It is the measure of how packed together the particles of that object are. An object placed in a liquid displaces a certain volume of the liquid, based on the relative density of the object and the liquid.
If an object is less dense than a liquid in which it is placed, it displaces a smaller volume of the liquid than its volume, hence only some part of the object will be seen to be under the liquid, the other part will float.
If an object is denser than the liquid in which it is placed, it displaces a larger volume of the liquid than its own volume, making the object to sink and is submerged, sometimes to the bottom of the liquid, but mostly below the point at which it was released.
Finally, if the density of an object and the liquid into which it is submerged is the same. the object's mass per unit volume is the same as the liquid's mass per unit volume, hence the weight and force created due to density will balance and cancel each other out hence making the object to remain where it was submerged.
Answer:
Explanation:
By Ohms Law, Voltage = Current * Resistance
Keeping the voltage the same and doubling the resistance, the current will be halved.
So the new current = 1.5/2 = 0.75A
Answer:
There are actually three, Kepler's laws that is, of planetary motion: 1) every planet's orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet's orbital period is proportional to the cube of the semi-major axis of its