Answer:
I=1.48 A
Explanation:
Given that
B=3.1 x 10⁻5 T
b= 4.2 cm
l= 9.5 cm
The relationship for magnetic field and current given as

Where

By putting the values


D=26.03 m⁻¹



I=1.48 A
Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s



While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

Substitute the given values to find "terminal speed"




The “terminal speed” of the ball bearing is 5.609 m/s
The answer is attract. Hope it helps! :)
Answer:
a = 17.68 m/s²
Explanation:
given,
length of the string, L = 0.8 m
angle made with vertical, θ = 61°
time to complete 1 rev, t = 1.25 s
radial acceleration = ?
first we have to calculate the radius of the circle
R = L sin θ
R = 0.8 x sin 61°
R = 0.7 m
now, calculating at the angular velocity


ω = 5.026 rad/s
now, radial acceleration
a = r ω²
a = 0.7 x 5.026²
a = 17.68 m/s²
hence, the radial acceleration of the ball is equal to 17.68 rad/s²