Vp = 155 V
R = 53 ω
(a) RMS emf output
Vrms = 0.7071Vp = 0.7071*155 = 109.6 V
(b) RMS current
From ohm's law,
Vrms = IrmsR => Irms = Vrms/R = 109.6/53 = 2.07 Amps
<span>Presuming that the gas behaves like an ideal gas:
Volume in cubic meter V= 59.6x 0.001= 0.0596 mÂł
n=3.01 mol
Temperature in Kelvin T=44.9+273.15=318.05 K
R= 8.413 J/ mol K
The ideal gas law:
PxV= nxRxT
Substituting the give information from the question into the ideal gas law:
P= 3.01 mol x 8.413 J/mol k x 318.05 K / 0.0596= 133544.392 Pa
1351.3458 Pa / 101325 Pa/ atm = 1.3179806 atm</span>
Answer:
The value is ![d = 0.000125 \ m](https://tex.z-dn.net/?f=d%20%3D%200.000125%20%5C%20%20m)
Explanation:
From the question we are told that
The wavelength is ![\lambda = 500 \ nm = 500 *10^{-9 } \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20500%20%5C%20%20nm%20%3D%20%20500%20%2A10%5E%7B-9%20%7D%20%5C%20%20m)
The distance is of the screen is ![D = 2.0 \ m](https://tex.z-dn.net/?f=D%20%3D%20%202.0%20%5C%20m)
The width of the fifth dark spot is ![y = 4.00 \ cm = 0.04 \ m](https://tex.z-dn.net/?f=y%20%20%3D%20%204.00%20%5C%20%20cm%20%20%3D%20%200.04%20%5C%20%20m)
Generally the width of a fringe is mathematically represented as
![y = \frac{ n * \lambda * D }{d }](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%20n%20%20%2A%20%20%5Clambda%20%20%2A%20%20D%20%7D%7Bd%20%7D)
=> ![0.04 = \frac{ 5 * 500 *10^{-9} * 2 }{d}](https://tex.z-dn.net/?f=0.04%20%20%3D%20%20%5Cfrac%7B%205%20%20%2A%20500%20%2A10%5E%7B-9%7D%20%20%2A%20%202%20%7D%7Bd%7D)
=> ![d = \frac{ 5 * 500 *10^{-9} * 2 }{0.04}](https://tex.z-dn.net/?f=d%20%3D%20%20%5Cfrac%7B%205%20%20%2A%20500%20%2A10%5E%7B-9%7D%20%20%2A%20%202%20%7D%7B0.04%7D)
=> ![d = 0.000125 \ m](https://tex.z-dn.net/?f=d%20%3D%200.000125%20%5C%20%20m)
Answer:
![s = v_0\sqrt{\frac{2\alpha}{g}}](https://tex.z-dn.net/?f=s%20%3D%20v_0%5Csqrt%7B%5Cfrac%7B2%5Calpha%7D%7Bg%7D%7D)
Explanation:
When the suitcase is being released, it has a horizontal velocity of v0, a vertical velocity of 0 and vertical acceleration of g
The time it takes for it to drop by the vertical distance of α under acceleration g can be solved using the following equation of motion
![\alpha = gt^2/2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20gt%5E2%2F2)
![t^2 = \frac{2\alpha}{g}](https://tex.z-dn.net/?f=t%5E2%20%3D%20%5Cfrac%7B2%5Calpha%7D%7Bg%7D)
![t = \sqrt{\frac{2\alpha}{g}}](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Calpha%7D%7Bg%7D%7D)
This is also the time it takes for the suitcase to travel horizontally at the rate of v0. So the horizontal distance it travels is
![s = v_0t = v_0\sqrt{\frac{2\alpha}{g}}](https://tex.z-dn.net/?f=s%20%3D%20v_0t%20%3D%20v_0%5Csqrt%7B%5Cfrac%7B2%5Calpha%7D%7Bg%7D%7D)
<span>The flywheel is solid cylindrical disc. Moment of inertial = ½ * mass * radius^2
Mass = 40.0 kg
Radius = ½ * 76.0 cm = 38 cm = 0.38 meter
Moment of inertial = ½ * 41 * 0.36^2
Convert rpm to radians/second
The distance of 1 revolution = 1 circumference = 2 * π * r
The number of radians/s in 1 revolution = 2 * π
1 minute = 60 seconds
1 revolution per minute = 2 * π radians / 60 seconds = π/30 rad/s
Initial angular velocity = 500 * π/30 = 16.667 * π rad/s
170 revolutions = 170 * 2 * π = 340 * π radians
The flywheel’s initial angular velocity = 16.667 * π rad/s. It decelerated at the rate of 1.071 rad/s^2 for 48.89 seconds.
θ = ωi * t + ½ * α * t^2
θ = 16.667 * π * 48.89 + ½ * -1.071 * 48.89^2
2559.9 - 1280
θ = 1280 radians</span>