A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn
1 answer:
Answer:
Electric field will remain same.
Explanation:
Given that
At initial condition ,charge q = +3μ C
Electric field E = 4106 N/c
As we know that
Electric field due to charge q
Now when charge is replaced by new charge q'= -3μ C
From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.
So we can say that electric field will remain same.
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Answer:
Explanation:
is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:
A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:
is the centripetal force and is defined as:
Here is the proton's speed and
is the radius of the circular motion. Replacing this in (1) and solving for r:
Recall that 1 J is equal to , so:
We can calculate from the kinetic energy of the proton:
Finally, we calculate the radius of the proton path:
Answer:
a). Maximum Length L=0.929m
b). T=0.83 Hz or 1.2s
c). Longer, the effortless waling T=2.1 Hz or t=0.475s
d). t=1.2s V=0.774
t=0.475s V=1.95
Explanation:
Length legs=L=1.1m
angle=50
the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image
Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length
a).
Length of the step
L=0.464m*2
L=0.928m
b).
period=T
c).
The period is the inverse of the time of the motion so, the T1 is faster that the T because
d).
The speed is the relation between the distance with time so: