Calculate the change in internal energy (δe) for a system that is giving off 25.0 kj of heat and is changing from 12.00 l to 6.0
2 answers:
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(1) A - B
(2) B - C
(3) - A + B - C
(4) 3A - 2C
(5) - 2A + 3B - C
(6) 2A - 3 (B - C)
Answer:
(1) (3,-5,-4)
(2) (-5, 4, 0)
(3) (-6, 4, 3)
(4) (-3, -2, -11)
(5) (-11, 14, 8)
(6) (17, -12, -6)
Explanation:
A⃗ =(1,0,−3)
B⃗ =(−2,5,1)
C⃗ =(3,1,1)
Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.
1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)
2) (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)
3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)
4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)
5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)
6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)
Answer:
26 days
Explanation:
m = 9.4×1021 kg
r= 1.5×108 m
F = 1.1×10^ 19 N
We know Fc =
==> 1.1 × = (9.4 ×
×
) ÷ 1.5 ×
==> 1.1 × =
× 6.26×
==> = 1.1 ×
÷ 6.26×
==> = 0.17571885 ×
==> v= 0.419188323 × m/sec
==> v= 419.188322834 m/s
Putting value of r and v from above in ;
T= 2πr ÷ v
==> T= 2×3.14×1.5× ÷ 0.419188323 ×
==> T = 22.472× 100000 = 2247200 sec
but
86400 sec = 1 day
==> 2247200 sec= 2247200 ÷ 86400 = 26 days