Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ =
= 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = 
0.005 = 2 x 3.142 x r
r = 0.005/6.284 =
= 0.0079 m =<em> 7.9 mm</em>
Answer:
They slow down.
Explanation:
The collided so the slow down till they stop.
It's Z.
Without any force acting on it an object travels in a straight line.
In order to bend away from a straight line the object needs a force acting on it.
In order to move along a circle, the force on the object points toward the center of the circle. It's called the centripetal force.
Since the object's direction is changing it has acceleration.
The acceleration points toward the center of the circle.
Answer: i would say true uwu
Explanation:<u><em>hope this helps! </em></u>
<u><em> ....Max- </em></u>
Answer:
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
Explanation:
Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.
Since there is a net force, there is an acceleration and thus an increasing velocity.
When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.
Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.
<u>So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.</u>
So, the answer is a.