The magnetic force between two wires is 0.052 N which is attract each other.
We need to know about magnetic force on a current-carrying wire formula to solve this problem. The magnetic force on two wires with same direction of current is
F = μ₀ . I1 . I2 . L / ( 2π . r )
where μ₀ is vacuum permeability (4π×10‾⁷ H/m) F is the magnetic force, I is current, L is the length of wire, r is distance of 2 wires.
From the question above, we know that:
L = 25 m
r = 6 cm
I1 = I2 = 25 A
By substituting the parameter, we get
F = μ₀ . I1 . I2 . L / ( 2π . r )
F = 4π×10‾⁷ . 25 . 25 . 25 / (2π . 0.06)
F = 0.052 N
Hence, the force between two wires is 0.052 N which is attract each other.
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kinetic <span>energy when the pitcher has thrown it or when one of the other players has thrown the ball. The baseball also has </span>kinetic<span> energy when the batter hits the ball. when you catch the ball it is potential energy.</span>
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
We first add the x and y components of the vector to get the x and y components of the resultant vector:
(0 - 1, 1 + 3)
-1 , 4
magnitude = √(1² + 4²)
magnitude = 4.12
The answer is B
