Answer:
v = √(10gh/7)
Explanation:
Initial gravitational energy = final rotational energy + kinetic energy
PE = RE + KE
mgh = ½ Iω² + ½ mv²
For a solid sphere, I = ⅖ mr².
For rolling without slipping, ω = v/r.
mgh = ½ (⅖ mr²) (v/r)² + ½ mv²
mgh = ⅕ mv² + ½ mv²
mgh = 7/10 mv²
10/7 gh = v²
v = √(10gh/7)
the answer is c. The induced current will increase
B is the correct answer . That make the most sense
Answer:
Explanation:
Given that the angular velocity is constant.
The actual weight is W = 549N
The ride apparent weight is
W' = 1.5N
Apparent weight at the top of the ride?
Using newton second law of motion
ΣF = m•ar
ar is the radial acceleration
N — W = —m•ar
N = W —m•ar
N = mg —m•ar
N = m(g—ar)
The apparent weight is equal to the normal
W'(top) = N = m(g—ar)
W'(top) = m(g—ar)
W'(top) = mg —m•ar
We know that the actual weight is
W=mg
Also, the apparent weight is
W' = m•ar
Then
W'(top) = Actual weight - apparent
W'(top) = 549—1.5
W'(top) = 547.5 N
AC stands for Alternating Current, according to physics terminology