Answer:
x = D (M/M-m) 2.41
Explanation:
a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation
F = g m₁m₂ / r²
Σ F = 0
F1- F2 = 0
F1 = F2
We set the reference system in the body of greatest mass (M) the planet
F1 = g m₁ M / x²
F2 = G m1 m / (D-x)²
G m₁ M / x² = G m₁ m / (D-x)²
M (D-x)² = m x²
MD² -2MD x + M x² = m x²
x² (M-m) -2MD x + MD² = 0
We solve the second degree equation
x = [2MD ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)
x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)
x = D {M ± Ra (2M²-mM)} / (M-m)
x = D (M ± M √ (2-m/M)) / (M-m)
x = D (M / (M-m)) (1 ±√ (2-m/M)
Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root
x = D (M / (M-m)) (1 + √2)
x = D (M/M-m) 2.41
b) X = 2/3 D
x = D (M/M-m) 2.41
2/3 D = D (M/(M-m)) 2.41
2/3 (M-m) = M 2.41
2/3 M - 2/3 m = 2.41 M
1.743 M = 0.667 m
M/m = 0.667/1.743
M/m = 0.38
Answer:
η = 2.57%
Explanation:
Given:
Output power of the steam power plant, P(out) = 200 MW
Consumption of coal, m = 700 tons/h = (700 × 10³)/3600 kg/s= 194.44 kg/s
Heating capacity of the coal, Cv = 40000 kJ/kg
now,
Input power, P(in) = mCv
or
P(in) = 194.44 kg/s × 40000 = 7777.77 MW
thus,
efficiency, η = P(out)/P(in) = (200 MW) / (7777.77 MW) = 0.02571
or
η = 2.57%
The definition for model system is for particular species of animal that has been developed over many years to be experimentally powerful to answer particular questions
Answer:
a=0.05m/s²
Explanation:
force=mass × acceleration
f=ma
f=50N
m=1000kg
a=?
50=1000×a
50=1000a
a=50/1000
a=0.05m/s²
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Answer:
P=38.0W
Explanation:
First find work done using
Ew=Fd
Ew=25×7.3
=182.5J
Then find power using
P=E/t
P=182.5/4.8
P=38.0W
