Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
Answer:
Ratio of length will be 
Explanation:
We have given time period of the pendulum when length is
is 
And when length is
time period 
We know that time period is given by

So
----eqn 1
And
-------eqn 2
Dividing eqn 2 by eqn 1

Squaring both side

Answer:
b)
Explanation:
By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.
This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.
(As the positive charge would move away from positive charges and would be attracted by negative ones).
So, the combination of answers that is true is b) (positive, negative, positive).
Answer:
Part a)
Induced EMF when length vector is along Z direction is 0.72 V
Part b)
Induced EMF when length vector is along Y direction is ZERO
Explanation:
As we know that the motional EMF induced in the wire is given as

1)
As we know that



now we have

so we have

2)
If the length vector is along Y direction then we have

so again we have

so we have
EMF = 0