Answer:
Answer was deleted first time, the answer is 917 N
Explanation:
Hope this helps!
Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a).
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
<span>Radius = 150 cramwells </span>
<span>t = sqrt(150/1.6) = 9.68 s.
I hope this helps.</span>
<h2>
Answer:53.63
</h2>
Explanation:
The equations of motion used in this question is 
When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.
But,due to gravity,the object accelerates downward at a rate of
.
In X-Direction,
Given that initial velocity=
=
Using
,

In Y-Direction,
Given that initial velocity=
=
Using
,



Force]/[force] = Newon/Newton = 1
Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

B is magnetic field

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.