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3 years ago

PLEASE HELP!!!How are middle latitude cyclones and tropical cyclones different from each other?

1 answer:

defon3 years ago

8 0

In contrast, extratropical cyclones have their strongest winds near the tropopause, which is about 8 miles above the surface. These differences are due to the tropical cyclone being “warm-core” in the troposphere, whereas extra-tropical cyclones are “warm-core” in the stratosphere and “cold-core” in the troposphere.

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What is a materials ability to be dissolved in a solvent

weeeeeb [17]

Answer: Solubility

Solubility is the ability to be dissolved. Saturation is when the concentration is too high(more than solubility) that when you add another material it won't dissolve. Solute is the material that dissolved. Solvent is the material that used for dissolving

5 0

3 years ago

An AC generator with a variable frequency and an RMS voltage output of 120 Volts is connected in series to a 3.1 μF Capacitor an

Maksim231197 [3]

Answer:

The RMS voltage across the resistor = 28 V

Explanation:

Capacitor: A capacitor is an electrical device that has the ability to store electrical charges in an electrical circuit. It is expressed in Farad (F)

Resistor: A resistor is an electrical device that oppose the flow of electric current in a circuit. It is expressed in ohms (Ω)

RMS Voltage : RMS voltage value of an alternating voltage is defined as that value of steady voltage which would dissipate heat at the same rate in a given resistance

Since the it is a series circuit, the total voltage is divided across the resistance and the capacitor.

Vt = V₁ + V₂...........................Equation 1

Where Vt = total Rms voltage = 120 V , V₁ = Rms voltage across the Capacitor = 92 V, V₂ = Rms voltage across the resistor.

Making V₂ the subject of the equation in equation 1 above,

V₂ = Vt - V₁ = 120 - 92

V₂ = 28 V.

The RMS voltage across the resistor = 28 V

7 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$