The solution to the questions are given as


- the direction of induced current will be Counterclock vise.
<h3>What is the direction of the
current induced in the loop, as viewed from above the loop.?</h3>
Given, $B(t)=(1.4 T) e^{-0.057 t}$




(b) 

c)
In conclusion, the direction of the induced current will be Counterclockwise.
Read more about current
brainly.com/question/13076734
#SPJ1
The question is incomplete but still I answer to assume your thinking.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
<span>∑ F = ma
</span>As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
<span>∑ F = 0</span>
N-wcosθ=0
N= m*g*cos25°
N= m*(9.8)*(0.9063)
N= 8.8817*mBy putting the value of mass(m)(not given in the question) you will get the answer.
Hopefully, this is the answer of your question.
Answer:
500ms times 2 would be when the ball reaches the max horizontal distance.
Then to find the angle, use the formula of time to reach max height t = u sin theta / g . With t being the max height time 500ms, u being 10m/s
For initial vertical velocity just use u sin theta.
Speed =distance over time