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Minchanka [31]
3 years ago
7

What kind of weather would MOST LIKELY lead to hurricane formation?

Physics
1 answer:
krok68 [10]3 years ago
7 0

Answer:

The recipe for a hurricane is a combination of warm, humid wind over tropical waters. The temperature of tropical waters must be at least 80 degrees F for up to 165 feet below the ocean’s surface. As this warm water meets the wind that blows west from Africa across the ocean, it causes the water to vaporize. The water vapor then rises into the atmosphere, where it cools and liquefies.

Explanation:

I hope this helps

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

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According to Kepler, what do all bodies in orbit around another have in common?
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The Coriolis effect _____.
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Read 2 more answers
g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t
jok3333 [9.3K]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}            (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

v_{max}=\omega A=2\pi f A      (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

a_{max}=\omega^2A=(2\pi f)^2 A

a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J

The total energy is 0.1J

(e) The displacement as a function of time is:

x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)

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