Answer:
A) Electric flux depends on the charge Q within and enclosed surface regardless of the shape of the surface - this concept is particularly useful for symmetric surfaces (centralized charge within a sphere) - field outside a uniform cylinder - whenever charges are symmetric with respect to the external environment.
B) Since the gravitational field has the same form as the electric field - F = G M1 M2 / R^2 the same arguments apply to the gravitational field - Field lines passing out thru an enclosed surface are sources of field whereas field lines passing inward thru an enclosed surface are sinks for the field
When is vertical velocity zero? Select all the apply
1. when an object changes direction.
2.the moment you kick a football in the air.
3. vertical velocity is zero initially when an object is dropped from a window.
4. the moment a ball rolls off a table.
5. the moment you throw a ball in the air.
Answer:
3
Explanation:
When the projectile reaches a vertical velocity of zero, this is the maximum height of the projectile and then gravity will take over and accelerate the object downward. The horizontal displacement of the projectile is called the range of the projectile, and depends on the initial velocity of the object.
Submarines change their depth by adding or removing air from rigid ballast tanks, thereby displacing seawater in the tanks. Consider a submarine that has a 700-m3 air-ballast tank originally partially filled with 100 m3 of air at 1650 kPa and 15°C. For the submarine to surface, air at 1650 kPa and 20°C is pumped into the ballast tank until it is entirely filled with air. The seawater leaves the tank at 15°C. Presume that air is added to the tank in such a way that the temperature and pressure of the air in the tank remain constant. Determine the final mass of the air in the ballast tank under this condition. Also determine the total heat transfer while the tank is being filled in this manner. The gas constant of air is R = 0. 287 kPa·m3/kg·K. The specific heats of air at room temperature are cP = 1. 005 kJ/kg·K and cv = 0. 718 kJ/kg·K. The specific volume of water is taken 0. 001 m3/kg
Total heat transfer while the tank is being filled in this manner, Q = ΔH = mCpΔT = 88,147 kJ
What is ideal gas law?Ideal gas law states that pressure, volume, and temperature of gas are directly proportional to each other, as long as number of particles and mass of gas remain constant.
PV = nRT
P is pressure, V is volume, n is number of moles of gas, R is gas constant, and T is temperature.
V = nRT/P
n = PV/RT
n = (1650 kPa)(100 m3)/(0.287 kPa·m3/kg·K)(288 K) = 603.5 kg
V = nRT/P = (603.5 kg)(0.287 kPa·m3/kg·K)(293 K)/(1650 kPa) = 105.5 m3
m = nM = (603.5 kg)(28.97 kg/kmol) = 17,486 kg
Q = ΔH = mCpΔT
ΔH is change in enthalpy, m is mass of air, Cp is specific heat at constant pressure, and ΔT is change in temperature.
ΔH = CpΔT
Q = ΔH = mCpΔT = (17,486 kg)(1.005 kJ/kg·K)(5 K) = 88,147 kJ
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If the diameter of the pie is ten inches, the approximate arc length of one slice of pie is select one.
The arc length of one slice of pie is either 5.24 inches for 6 slices or 3.93 inches for 8 slices.
The arc length of one slice of pie can be found by dividing the circumference of the pie by the number of slices it is cut into.
The circumference of the pie can be found using the formula
C = πd
Where d is the diameter of the pie.
Substituting d = 10 inches,
we get:
C = π × 10 inches ≈ 31.42 inches
If we assume that the pie is cut into n equal slices, then the arc length of one slice is approximately:
Arc length ≈ C/n
If we want an approximate value for the arc length, we can use a value of n that is easy to work with,
such as n = 6 for 6 slices or n = 8 for 8 slices.
For n = 6, the arc length of one slice is approximately:
Arc length ≈ 31.42 inches / 6 ≈ 5.24 inches
For n = 8, the arc length of one slice is approximately:
Arc length ≈ 31.42 inches / 8 ≈ 3.93 inches
Therefore, the approximate arc length of one slice of pie is either 5.24 inches for 6 slices or 3.93 inches for 8 slices.
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A light wave has a frequency of 5.4*10^14 cycles per second and a wavelength of 5.5*10^-17 meter. What is the approximate speed of the wave?
A light wave has a frequency of 5.4 x 10⁻¹⁴cycles per second and a wavelength of 5.5 x 10⁻¹⁷ meter so the speed of the wave is 29.7 x 10⁻³m/s.
How to find the speed of the wave?v=λf, or velocity = wavelength x frequency, can be used to calculate a wave's speed. The distance a wave covers in a certain amount of time, such as the number of meters it covers every second, is known as its wave speed.
The formulae of the speed of the wave are,
v=λf, or velocity = wavelength x frequency
Frequency = 5.4 x 10⁻¹⁴ hertz
Wavelength = 5.5 x 10⁻¹⁷ m
v = 5.5 x 10⁻¹⁷x 5.4 x 10⁻¹⁴
= 29.7 x 10⁻³ m/s
Therefore, the speed of the wave is 29.7 x 10⁻³m/s.
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an object or is placed close to a thin converging lens. The diagram represents three way from the top of all passing through that lens.
Which option is correct A, B, C or D?
The type of image produced by the converging lens when object O is at that position will be virtual and enlarged, Option D.
What is a virtual and enlarged image?A virtual image is an image that is formed when light rays do not actually converge at a single point, but instead appear to converge when they are extended backwards. A virtual image is typically created by objects that are behind a lens or mirror that diverts the light rays in such a way that they appear to come from a different location.
An enlarged image, on the other hand, refers to a visual representation that has been magnified or made larger than the original size of the object being viewed.
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