Homosapiens I think? I’m not sure but I’m 30% sure
The formula is used when the statement of second law of motion is applied to the problem. If the statement is applied to the problem, The fomula will be used. If not, use the formula
<span>It will rotate with a greater angular velocity when it is travelling slowly. When the front of the car is just over the edge, the car will not rotate as the center of mass is still above ground, thus negates gravity's torque.
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Answer:
The ball was in air for 3.896 s
Explanation:
given,
g = 9.8 m/s², acceleration due to gravity,
If the launch angle is 45°, the horizontal range will be maximum.
The horizontal and vertical launch velocities are equal, and each is equal to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The time to attain maximum height is one half of the time of flight.
v = u + at ∵ v = 0 (max. height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The time of flight is twice of the maximum height time
2 t₁ = 3.896 s
The horizontal distance traveled is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in air for 3.896 s
Biot-Savart Law - Magnetic field around a straight conductor carrying a current I.
B = μ_0 I/2πr
where:
B is strength of magnetic field in T.
μ_0 is permeability of free space = 1.25664 x 10^ -6 T-m/A.
I is current in A.
r is distance from conductor in m.
<span>So I = 2πrB /μ_0</span>