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# What is the angle of dip at a place where the horizontal component and vertical component of the earth magnetic field are equal?

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## Related Questions

###### Asked in Physics, Earth Sciences

### What is angle between magnetic field and gravitational field of the earth?

That angle changes with location on the Earth. The only places
where the forces
of both of them are parallel would be if you were standing at
the Earth's north or
south magnetic poles.
The force of gravity is always nominally toward the center of
the Earth (local "down"),
with no horizontal component. But the force on a magnet due to
the Earth's magnetic
field has a horizontal component everywhere, except at the
magnetic poles.

###### Asked in Geometry

### Does diagonal mean an angle cut?

###### Asked in Physics

### Describe the horizontal component of a projectile motion?

I will try to describe in a very simple way, cuz I don't know it
any other way. Take the said vector magnitude times the cosine of
the angle from horizontal. In a two dimensional plane, say Joe
throws a football at an angle of 10 degrees from horizontal, at 100
miles an hour. So, the horizontal component will be = 100 x
cos(10)=98.48 miles an hour. The vertical component will be 100 x
sin(10) = 100 x cos(80)=17.36 miles an hour. The angles are in
degrees.

###### Asked in Physics

### Shell of mass 50 kg is fired at 60 degrees to the horizontal with the speed of 200ms Neglecting air resistance what is the kinetic energy of the shell in joules at its highest point or maximum height?

Well it is a question of Projetile Motion. Whenever you project
an object at some angle with horizontal it traces a curved path.
Suppose 'x' be the angle with horizontal, 'm' be its mass, 'u' be
its initial velocity. You might be knowing that any vector can be
resolved into horizonal and vertical component. Here horizontal
component of velocity will be ucosx and vertical will be usinx. At
highest point the vertical component of the velocity will become
zero. Hence it will have no velocity in vertically upward
direction. It will only have velocity in horizontal direction which
will be 'ucosx'.(The horizontal component of initial velocity will
not affected by force of gravity and hence it will be constant
throughout the motion.) So kinetic energy at highest point is:
KE=0.5*m*(ucosx)^2 KE=0.5*50*(200*cos60)^2=25*10000=250000 J