Answer:
one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/
Explanation:
| Impedance | = √ [R² +(ωL)²]
R² = 6800² = 4.624 x 10⁷
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²
= 2.0884 x 10⁻⁴ f²
| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵
(1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)
(2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²
Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴
= 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴
= 1.224 x 10¹⁴
= 122,400 GHz <== my calculation
11.1 MHz <== online impedance calculator
Obviously, I must have picked up some rounding errors
in the course of my calculation.
<span>Maritime tropical air masses develop over warm waters present in the tropics and Gulf of Mexico, where heat and moisture are carried to to the overlying air from the water below.
</span><span>
</span><span> Tropical air masses having northward movement carry warm moist air into the United States, thus increasing the potential for condensation. Generally the southern states experience tropical air masses. But, in winter season, southerly winds ahead of migrating cyclones <span>sometimes transport tropical air mass towards north.
</span></span><span><span>
</span></span><span><span>The counterclockwise winds related to northern hemisphere mid latitude cyclones play an important role in the movement air masses, carrying warm moist air towards north ahead of a low while dragging colder and drier air towards south.</span></span>
The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.
Answer:
F = 351×10³lb
Explanation:
Given the density
ρg = 64.6lb/ft³
Diameter d = 12ft
The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft
The pressure in the tank is
P = ρgh = 64.6 × 12 = 775.2lb/ft²
The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²
F = 351×10³lb.
