Answer: (a) Magnetic flux before rotation is 7.2 × 10^-8 Wb and magnetic flux after rotation is 0
(b) The average emf induced in the coil is 3.6×10^-4 V
Step-by-step explanation:
Here is the complete question:
In a physics laboratory experiment, a coil with 200 turns en-closing an area of 12 cm2 is rotated in 0.040 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 6.0×10^−5T. (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?
Step-by-step explanation: Please see the attachments below
You would do 7/8 x 8/1 which is 56/8 which simplifys to 7 so your answer would be 7
A :-) 1.) Given - base = 9 cm
height ( alt ) = 12 cm
hypotenuse ( hypo ) = x
Solution -
By Pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 9 )^2 + ( 12 ) ^2
( x )^2 = 81 + 144
( x )^2 = 225
( x ) = _/225
( x ) = 15 cm
.:. The value of x ( hypotenuse ) = 15 cm
2.) Given - base = 10 cm
Height = 24 cm
Hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 10 )^2 + ( 24 )^2
( x )^2 = 100 + 576
( x )^2 = 676
( x ) = _/676
( x ) = 26
.:. The value of x ( hypotenuse ) = 26 cm
3.) Given - base = 3 cm
Height = 7 cm
Hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 3 )^2 + ( 7 )^2
( x )^2 = 9 + 49
( x )^2 = 58
( x ) = _/58
( x ) = 7.6
.:. The value of x ( hypotenuse ) = 7.6 cm
4.) Given - base = 10 cm
Height = 6 cm
Hypotenuse = x
Solution -
By pythagorus theorem
( Hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 10 )^2 + ( 6 )^2
( x )^2 = 100 + 36
( x )^2 = 136
( x ) = _/136
( x ) = 11.6
.:. The value of x ( hypotenuse ) = 11.6 cm
5.) Given - hypotenuse = 24 cm
height = 6 cm
Base = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( 24 )^2 = ( x )^2 + ( 6 )^2
( x )^2 = ( 6 )^2 - ( 24 )^2
( x )^2 = 36 - 576
( x )^2 = -540
( x ) = _/-540
( x ) = 23.2
.:. The value of x ( base ) = 23.2 cm
6.) Given - base = 1 cm
height = 1 cm
hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 1 )^2 + ( 1 )^2
( x )^2 = 1 + 1
( x )^2 = 2
( x ) = _/2
( x ) = 1.4
.:. The value of x ( hypotenuse ) = 1.4 cm
7.) Given - hypotenuse = 21 cm
height = 8 cm
Base = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( 21 )^2 = ( x )^2 + ( 8 )^2
441 = ( x )^2 + 64
( x )^2 = 64 - 441
( x )^2 = -377
( x ) = _/-377
( x ) = 19.4
.:. The value of x ( base ) = 19.4
8.) given - height = 24 cm
Hypotenuse = 30cm
Base = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( 30 )^2 = ( x )^2 + ( 24 )^2
900 = ( x )^2 + 576
( x )^2 = 576 - 900
( x )^2 = -324
( x ) = _/-324
( x ) = 18
.:. The value of x ( base ) = 18 cm
9.) ( i ) lets find ‘x’
Given - base = 9 cm
height = 5 cm
hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 9 )^2 + ( 5 )^2
( x )^2 = 81 +25
( x )^2 = 106
( x ) = _/106
( x ) = 10.2
.:. The value of x ( hypotenuse )
= 10.2 cm
( ii ) lets find ‘y’
Given - base = 3 cm
height = 5 cm
Hypotenuse = y
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( y )^2 = ( 3 )^2 + ( 5 )^2
( y )^2 = 9 + 25
( y )^2 = 34
( y ) = _/34
( y ) = 5.8
.:. The value of y ( hypotenuse )
= 5.8 cm
The derivative of the function csc (5x) can be determined by following the trigonometric rule of differentiation. In this case, the derivative of csc x is -csc x cot x while that of 5x is 5. The total derivative and the answer to the problem asking for the is <span>derivative of csc(5x) is </span>-5 csc<span> 5x cot 5x. </span>
The answer is y - 3 = -1/2 (x - 2).
First, let's find the slope.
- m = y₂ - y₁ / x₂ - x₁
- m = 3 - 6 / 2 + 4
- m = -3/6
- m = -1/2
Then, the point slope form will be :
