Answer:
the maximum current is 500 A
Explanation:
Given the data in the question;
the B field magnitude on the surface of the wire is;
B = μ₀i / 2πr
we are to determine the maximum current so we rearrange to find i
B2πr = μ₀i
i = B2πr / μ₀
given that;
diameter d = 2 mm = 0.002 m
radius = 0.002 / 2 = 0.001 m
B = 0.100 T
we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A
so we substitute
i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷
i = 500 A
Therefore, the maximum current is 500 A
H = 11 m, the vertical distance that the bullet falls.
Initial vertical velocity = 0
Horizontal velocity = 144.7 m/s
The time, t, taken to fall 11 m is given by
(1/2)*(9.8 m/s²)*(t s)² = (11 m)
4.9t² = 11
t = 1.4983 s
If aerodynamic resistance is ignored, the horizontal distance traveled before the bullet hits the ground is
d = (144.7 m/s)*(1.4983 s) = 216.8 m
Answer: 216.8 m
Answer:
C) Inversely to its volume
Explanation:
As the volume decreases (Container gets smaller) the pressure increases. Conversely, If the volume increases (Container gets larger) the pressure decreases.
Player A needs the least amount of energy. The ball is light weight and she is closest to the goal so the momentum need to kick the ball will be the least and the distance is has to travel is the shortest. But player C needs the most amount of energy. The ball is heavy so it will take the most momentum to move the ball and over such a long distance. Hope this help idrk.
Hello =D
This problem is about cinematic
So
V = 45 mi/h
t = 2 h
Then
V= X/t
X = V*t
Then
X = (45)*(2)
X = 90 mi
Best regards
