Answer:
4050 gallons
Explanation:
In first half and hour both pumps are working property . There for it give, 45
gallons per minute . Half and hour is equal to 30 minutes.After one pump was broken only one pump is working 30 minutes .
For 1st 30 minutes ,( 2 pumps are working )
Gallons = 45 * 30 *2 = 2700 gallons
2nd 30 minutes ( 1 pump is working )
Gallons = 45*30*1 = 1350 gallons
Total gallons = 2700+1350
= 4050 gallons
Answer:
Approximately 66,8kg
Hope this help :D
Explanation:
F=mg. So 655N= 9.8 m/s² × mass.
Mass = 655 / 9.8
Answer:
v_f = 6.92 x 10^(4) m/s
Explanation:
From conservation of energy,
E = (1/2)mv² - GmM/r
Where M is mass of sun
Thus,
E_i = E_f will give;
(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)
m will cancel out to give ;
(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)
Let's make v_f the subject;
v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]
G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²
Mass of sun is 1.9891 x 10^(30) kg
v_i = 2.1×10⁴ m/s
r_i = 2.5 × 10^(11) m
r_f = 4.9 × 10^(10) m
Plugging in all these values, we have;
v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12
v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]
v_f = √[(441000000) + (435.38 x 10^(7))
v_f = 6.92 x 10^(4) m/s
Answer:
1 μF
Explanation:
To obtain the answer to the question, all we need to do is to calculate the equivalent capacitance of the capacitors. This can be obtained as illustrated below.
From the question given above, the following data were obtained:
Capacitor 1 (C₁) = 2 μF
Capacitor 2 (C₂) = 4 μF
Capacitor 3 (C₃) = 4 μF
Equivalent capacitance (Cₑq) =?
Cₑq = 1/C₁ + 1/C₂ + 1/C₃
Cₑq = 1/2 + 1/4 + 1/4
Cₑq = (2 + 1 + 1)/4
Cₑq = 4/4
Cₑq = 1 μF
Thus, the answer to the question is 1 μF
