Two concentric current loops lie in the same plane. The outer loop has twice the diameter of the inner loop. The inner loop carr
ies a 1.0 A current in a clockwise direction. What current, and which direction, should the outer loop carry such that the magnetic field in the center of the loops is zero?
1 answer:
You might be interested in
Answer:
a = 7.35 ft / s²
Explanation:
For this exercise we must use the kinematics relations
x = v₀ t + ½ a t²
as the runner leaves the starting line his initial velocity is zero
x = ½ a t²
a =
let's reduce the distance to foot
x = 60 yd (3ft / 1yd) = 180 ft
let's calculate
a = 2 180 / 7²
a = 7.35 ft / s²
Answer:
magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Explanation:
Given the data in the question;
Speed of carousel N = 24 rpm
From the diagram below, selected path direction defines the Axis of slip.
Hence, The Coriolis is acting along the axis of transmission
Now, we determine the angular speed ω of the carousel.
ω = 2πN / 60
we substitute in the value of N
ω = (2π × 24) / 60
ω = 2.5133 rad/s
Next, we convert the given velocity from mph to ft/s
we know that; 1 mph = 1.4667 ft/s
so
= 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s
Now, we determine the magnitude of the Coriolis acceleration
= 2(
× ω )
we substitute
= 2( 8.8002 ft/s × 2.5133 rad/s )
= 44.235 ft/s²
Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Answer:
statement B is true
Explanation:
since same force is applied by the compressed spring on both masses so their rate of change of momenta must be same and since the lighter block has lesser mass so it must have greater velocity to have an equal change in momentum as of heavier mass.
By relation:
,
comparing momenta of above two equations we get
KElighter (2) = KEheavier (4)
KElighter = 2 KEheavier