Answer:
the total force vector, on test charge is points from origin to point C( 1, 1 )
Explanation:
Given the data in the question, as illustrated in the image below;
from the Image, OA = 1, OB = AC = 1
so using Pythagoras theorem
a² = b² + c²
a = √( b² + c² )
so
OC = √( OB² + AC² )
we substitute
OC = √( OA² + AC² )
OC = √( 1² + 1² )
OC = √( 1 + 1 )
OC = √2
Coordinate of C( 1, 1 )
Hence, the total force vector, on test charge is points from origin to point C( 1, 1 )
Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.
Using Sine rule to solve this question
Sine rule => SinA/a = SinB/b = SinC/c = constant
The magnitude of wind is 50 with an angle of 60 degrees.
The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.
One side is 50 km/hr at an angle of 60 degrees.
sin 60°/200 = sin θ / 50
50 × sin 60° = 200 × sin θ
√3/2 = 4 × sin θ
√3/8 = sin θ
sin θ = 0.2165
θ = sin⁻¹(0.2165)
θ = 12.5°
So Lindsay have to fly the plane in the direction of [W 12.5° S].
Learn more about Sine Rule here:
brainly.com/question/27174058
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Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
The answer is A. When the the temperature increases the kinetic energy increases
Answer:
Explanation:
As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time
So we can say
now we have
here since magnetic field is constant so we have
now we have
now we have
