Below are the choices that can be found elsewhere:
A. (4.9 × 10-14 newtons) · tan(30°)
<span>B. (4.9 × 10-14 newtons) · sin(30°) </span>
<span>C. (4.9 × 10-14 newtons) · cos(30°) </span>
<span>D. (4.9 × 10-14 newtons) · arctan(30°) </span>
<span>E. (4.9 × 10-14 newtons) · arccos(30°)
</span>
<span>Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)</span>
Answer:
It is most likely option A B and C
Answer:
P = 7.28 N.s
Explanation:
given,
initial momentum of cue ball in x- direction,P₁ = 9 N.s
momentum of nine ball in x- direction, P₂ = 2 N.s
momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s
momentum of the cue after collision = ?
using conservation of momentum
in x- direction
P₁ + p = x + P₂
p is the initial momentum of the nine balls which is equal to zero.
9 + 0 = x + 2
x = 7 N.s
momentum in x-direction.
equating along y-direction
P'₁ + p = y + P'₂
0 + 0 = y + 2
y = -2 N.s
the momentum of the cue ball after collision is equal to resultant of the momentum .


P = 7.28 N.s
the momentum of the cue ball after collision is equal to P = 7.28 N.s
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
The focal length (in meters) of a lens whose radius of curvature is 9. 2 m and has a refractive index 1.2 will be 18.4 m
The focal length of a lens is determined when the lens is focused at infinity. Lens focal length tells us the angle of view—how much of the scene will be captured—and the magnification—how large individual elements will be.
Focal length of a lens is the distance between center of lens and focal point . Focal point is a point on principal axis , at which light rays parallel to principal axis meet after refraction through lens or seem to meet after refraction .
The radius of curvature is the radius of sphere formed by the convex or concave mirror. It is also equal to the distance between the pole and center of curvature. The sign convention for focal length and radius of curvature is the same.
focal length = 2 * radius of curvature
given
radius of curvature = 9.2 m
focal length = 2 * 9.2
= 18.4 m
To learn more about focal length here
brainly.com/question/16188698
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