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1. ### If a 20 kg of mass, m, is attached to the above spring in problem 20 (spring constant k), what will the frequency of oscillation, f, be?

f = 1/(2 * π) * √(k/m) The first step is solve problem 20 to determine the spring constant... to determine the increase of the length. ∆ L = 0.15 – 0.10 = 0.05 meter k = 4 ÷ 0.05 = 80 N/m Now we can use the spring constant to calculate the...

2 Answers · Science & Mathematics · 16/08/2019

2. ### Attach an object A of mass m to the left end of the spring with a spring constant k and an object B of 2m to the right end as shown?

...2 ] + 0.5k[y(t) - x(t) - a]^2 = 0.5m(v_0)^2 (v_0)^2 - (dx/dt)^2 - 2(dy/dt)^2 = (k/m)[y(t) - x(t) - a]^2 4. When dx/dt = 0 then dy/dt = v_0 / 2 Plug...

1 Answers · Science & Mathematics · 28/06/2019

3. ### In the figure, the total resistance is 12.4 kΩ , and the battery's emf is 29.2 V . If the time constant is measured to be 25.2 μs .?

...2*10^-6/(12.4*10^3) = 2.032*10^-9 F (2.032 nF) 17.2/29.2 = e^-k e^-k = 0.589 -k*ln e = ln 0.589 ..(ln e = 1) k = 0.5293 = t/Ƭ t = 0.5293...

2 Answers · Science & Mathematics · 13/03/2019

4. ### .210 moles of a diatomic gas are in a piston the gas dos 185J of work while 390 of heat are added what is the change in temperature?

For diatomic ideal gas U = 5/2 nRT (should be Delta on U and T) U = Q - W T = (Q-W)/(5/2 nR) = (390-185)/(5/2 0.21*3.814)

1 Answers · Science & Mathematics · 23/05/2019

5. ### If an Aircraft fly at an altitude of 30,000 feet what is air pressure and air density in {kg\m-3}, air temperature in (k).?

...m, 9.1 km pressure is about 30 kPa Temp is about 230 K density is about 0.46 kg/m³"{kg\m-3}" is meaningless...

3 Answers · Science & Mathematics · 02/11/2019

6. ### If an Aircraft fly at an altitude of 30000 how can I calculate air temperature in (k) air pressure and air density in (kg/m'3) according to ?

...m, 9.1 km pressure is about 30 kPa Temp is about 230 K density is about 0.46 kg/m³ how many times are you...

1 Answers · Science & Mathematics · 04/11/2019

7. ### What is the resultant pressure if 0.7 mol ofideal gas at 273 K and 2.37 atm in a closedcontainer of constant volume is heated?

Let’s use the following equation to determine the final pressure of the gas. P1 ÷ T1 = P2 ÷ T2 2.37 ÷ 273 = P2 ÷ 610 P2 * 273 = 1,445.7 P2 =1,445.7 ÷ 273 This is approximately 5.3 atm. I hope this is helpful for you.

2 Answers · Science & Mathematics · 12/05/2019

8. ### a m=1.50 kg block attached to a massless spring (k=1400N/m). The block is pulled to a position 5.00 cm to the left of its equilibrium?

... energy is conserved PE max = PE at 2cm + KE at 2cm 1/2(1400)(0.0500^2) = 1/2(1400)(0.0200^2) + 1/2 (1.50) v^2 1/2 's cancel 3.5 = 0.56 + 1.50 v^2 1.50 ^2 = 2.94 v^2 = 1.96 v = 1.4 m/s When you get a good response, please consider giving a best answer. This...

1 Answers · Science & Mathematics · 17/04/2019

9. ### Two 4.4 μF capacitors, two 2.5 kΩ resistors, and a 11.4 V source are connected in series.?

two 4.4 in series are equal to 2.2 two 2.5k in series are equal to 5 time constant = 2.2µ x 5k = 11 ms voltage across 5k is 1.5•5 = 7.5 volts voltage across cap = 11.4–7.5 = 3.9 volts v = v₀[1–e^(–t/τ)] 3.9 = 11.4[1–e^(–t/11m)] 1–e^(–t/11m) = 0.3421 e^(–t/11m) = 0.6579 –t/11m...

1 Answers · Science & Mathematics · 02/05/2019

10. ### a spring with k=33.5 n/m has 1.2kg mass attached it is pulled .12m and released how much mechanical energy does it have?

Assuming the system has no friction, it's a simple harmonic oscillator and therefore has constant mechanical energy. When the spring is fully stretched, all of that mechanical energy is spring potential E = PS E = ½kA²...

2 Answers · Science & Mathematics · 28/05/2019