A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
1 answer:
Answer:
Magnetic field will be ZERO at the given position
Explanation:
As we know that the magnetic field due to moving charge is given as
so here we know that for the direction of magnetic field we will use
so we have
so magnetic field must be ZERO
So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero
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Answer:
x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components
1) Focal length
We can find the focal length of the mirror by using the mirror equation:
(1)
where
f is the focal length
is the distance of the object from the mirror
is the distance of the image from the mirror
In this case,
, while 
(the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
from which we find
2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
We can find the focal length of the mirror by using the mirror equation:
where
f is the focal length
In this case,
from which we find
2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is: