The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
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Answer:
A. 50.7 J
B. 33.18 J
Explanation:
At the start of the motion the kinetic energy of the ball would be
When the ball gets to the highest point, vertical velocity must be 0, only horizontal velocity contributes to the kinetic energy. Since air resistant can be neglected, horizontal energy is preserved since the start
So the kinetic energy at this point is
Answer:
magnetic flux ΦB = 0.450324 × weber
current I = 1.02484 A
Explanation:
Given data
length a = 2.2 cm = 0.022 m
width b = 0.80 cm = 0.008 m
Resistance R = 0.40 ohms
current I = 4.7 A
speed v = 3.2 mm/s = 0.0032 m/s
distance r = 1.5 b = 1.5 (0.008) = 0.012
to find out
magnitude of magnetic flux and the current induced
solution
we will find magnitude of magnetic flux thorough this formula that is
ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]
here μ is 4π × put all value
ΦB = (4π × 4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]
ΦB = 0.450324 × weber
and
current induced is
current = ε / R
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
put all value
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
current = 4π × (4.7) (0.022) (0.008) (0.0032) / 2π(0.40) [(0.012² ) - (0.008/2 )² ]
current = 1.02484 A