Answer: Gradient Wind
Explanation:
Gradient wind, is the wind that accounts for air flow along a curved trajectory. It is an extension of the concept of geostrophic wind; for example the wind assumed to move along straight and parallel isobars (lines of equal pressure). The gradient wind represents the actual wind better than the geostrophic wind, especially when both wind speed and trajectory curvature are large, because they are in hurricanes and jet streams.
Answer:
yes ............................................ks
Explanation: is good
Answer:
density of cube =11.605 g/cm³
Explanation:
density of a substance is the mass per unit volume of that substance.
the density of a substance = 
volume of a cube = l³,
l = 19.0mm , lets convert mm to cm
1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm
length of cube =1.9cm
volume of cube = 1.9³
density of cube = 
density of cube =11.605 g/cm³
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
Hence, The maximum height above the point of release is 11.653 m.
The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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