Answer:
i = 4.9 A
Explanation:
The expression for the magnetic force in a wire carrying a current is
F = i L x B
bold letters indicate vectors.
The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition
F - W = 0
i L B = m g
They indicate the linear density of the cable λ = 0.2 kg / m
λ = m / L
m = λ L
we substitute
i B = λ g
i = 
let's calculate
i = 0.2 9.8 / 0.4
i = 4.9 A
Answer:
b) 1.67×10^7 m/s
Explanation:
The solution is attached in the attachment section
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds).
Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.
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The area under the velocity time graph is 125 m and the meaning of the area is displacement.
<h3>
What is area under velocity - time graph?</h3>
The area under a velocity time graph represents the displacement of the object.
total area of the graph = A1 + A2
total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)
total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)
total area of the graph = 125 m
Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.
Learn more about velocity time graph here: brainly.com/question/4710544
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By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
