Why not two magnetic field lines can intersect? why not it is possible?

1 answer:

4 0

At each point on a 'line', the direction of the 'line' is the direction of the force
on a small test magnet placed in the field at that point.

If two 'lines' crossed at the same point, that means a small test magnet placed
at that point in the field would feel a force in two different directions.

But even if that were true, then the net effect on the small test magnet would be
the vector sum of the two forces, and they would be represented by a single net
force anyway, and therefore by a single field 'line' at that point.

You might be interested in

the radius of the tires on a particular vehicle 0.62m if the tires are rotating 5 times per second, what is the velocity of the

tankabanditka [31]

Circumference of the tire = (2 pi) x (radius)

   = (2 pi) x (0.62 meter)

   = 3.9 meters

If the tire never slips or skids, then the speed of the vehicle is

   speed = (distance) / (time to cover the distance)

   = (5 x 3.9 meters) / 1 second

   = 19.48 meters/second .

(about 43.6 miles per hour) .

We can't say anything about the vehicle's velocity, because we have
no information about the direction in which it's heading.

8 0

3 years ago

State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati

juin [17]

F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)

6 0

3 years ago

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$