Electrons move in atomic orbitals (or subshells). there are four different orbital shapes (s p d f). in each shell, the s subshell is at a lower energy than the p. an orbital diagram is used to determine an atom's electron configurations
Your answer is B
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Answer:
ΔU = 5.21 × 10^(10) J
Explanation:
We are given;
Mass of object; m = 1040 kg
To solve this, we will use the formula for potential energy which is;
U = -GMm/r
But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.
Thus;
ΔU = -GMm((1/r_f) - (1/r_i))
Where;
M is mass of earth = 5.98 × 10^(24) kg
r_f is final radius
r_i is initial radius
G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Since, it's moving to altitude four times the Earth's radius, it means that;
r_i = R_e
r_f = R_e + 4R_e = 5R_e
Where R_e is radius of earth = 6371 × 10³ m
Thus;
ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)
× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))
ΔU = 5.21 × 10^(10) J
The answer to your question is A.
Answer:
a) 166.4 s
b) (2.155 × 10⁷) s
Explanation:
15600 KWh for a year,
1 year consists of 365 × 24 hours = 8760 hours.
So, the power consumed in a year for an average household = (Energy/time)
= (15600/8760) = 1.781 KW = 1781 W
a) If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take
The power required for this lifting = (mgh/t)
m = 1800 kg
g = 9.8 m/s²
h = 16.8 m
t = ?
P = 1781 W
1781 = (1800×9.8×16.8)/t
t = (296,352/1781)
t = 166.4 s
b) how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km
The power required for this lifting = (mgh/t)
m = 405000 kg
g = 9.8 m/s²
h = 9.67 km = 9670 m
t = ?
P = 1781 W
1781 = (405000×9.8×9670)/t
t = (38,380,230,000/1781)
t = 21,549,820 s = (2.155 × 10⁷) s
Hope this Helps!!!
