Which of the following classifications of star temperature is hottest? K O B M
2 answers:
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Explanation:
It is given that,
Mass of the ball, m = 0.06 kg
Initial speed of the ball, u = 50.4 m/s
Final speed of the ball, v = -37 m/s (As it returns)
(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :
J = -5.24 kg-m/s
(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.
W = -35.1348 Joules
Hence, this is the required solution.
Answer
given,
mass of base ball = 0.14 kg
speed before it made the contact with the ball (V i) = 42 m/s
speed after batter hit the ball(V f) = - 48 m/s
a)
impulse = change in momentum
=
=
= -12.6 Kg m/s
Magnitude of impulse = 12.6 Kg m/s
b)
Force =
=
Force = 2520 N
Answer:
current in loops is 52.73 μA
Explanation:
given data
side of square a = b = 2.40 cm = 0.024 m
resistance R = 1.20×10^−2 Ω
edge of the loop c = 1.20 cm = 0.012 m
rate of current = 120 A/s
to find out
current in the loop
solution
we know current formula that is
current = voltage / resistance .................a
so current = 1/R × d∅/dt
and we know here that
flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b
so
d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c
so from equation a we get here current
current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt
current = ( 4π××0.024 / 2π(1.20×
) × ln (0.024 + 0.012/0.012) × 120
solve it and we get current that is
current = 4 ×× 1.09861 × 120
current = 52.73 × A
so here current in loops is 52.73 μA