Calculate the impulse of a 0.140 kg baseball that starts at rest on a tee and reaches a final velocity of 20 m/s after being str
2 answers:
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Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 × m
diameter d = 0.074 cm = 74 × m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W
Answer:
a)-1.014x J
b)3.296 x J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x
x 47 x 110/ 3.4 => -1.014x
J
b) at d= 0.8m (3.4-2.6) and =-1.014x
J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+
=
+
As sphere starts from rest and sphere A is fixed at its place, therefore is zero
=
+
The final potential energy is
= - GMaMb/d
Solving for ' '
=
+ GMaMb/d => -1.014x
+ 6.67 x
x 47 x 110/ 0.8
= 3.296 x
J