Dubai this part of india has a lot of oil is consider one that has more oil wells.
Answer:
C. 28.09 amu
Explanation:
The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abundance) and X-30 (29.974 amu, 3.10% abundance).
The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.
The atomic weight is computed as follows:
atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of X-30 × fractional abundance
atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 × 0.0310
atomic weight = 25.8031871 + 1.3531792 + 0.929194
atomic weight = 28.0855603 amu
To 2 decimal place atomic weight = 28.09 amu
Answer:
1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x 
J
2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x J
Explanation:
given information:
= 3 nC = 3 x 
C
= 2 nC = 2 x C
r = 50 cm = 0.5 m
the electric potential energy of the electron when it is at the midpoint
potential energy of the charge, F
F = k 
where
k = constant (8.99 x 
)
electron charge, 
= - 1.6 x 
C
since it is measured at the midpoint,
r = 
= 0.25 m
thus,
F = 
= k
+ k
= 
(
)
= (8.99 x 
)( - 1.6 x )(3 x +2 x )/0.25
= - 2.9 x J
the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge
= 10 cm = 0.1 m
= 0.5 - 0.1 = 0.4 m
F = k + k
= 
(
+
)
= (8.99 x )( - 1.6 x )(3 x /0.1+2 x /0.4)
= - 5.04 x J
His power output was 3 Watt (360 Joule/120 seconds). The power output can be calculated by dividing the quantity of work by the amount of second needed for the activity and also by multiplying the force amount with the velocity of the activity. The power output usually used for measuring the ability of machine for doing its job.
