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3 years ago

10

Which term best describes a severe weather event characterized by heavy snowfall, strong winds, and reduced visibility?

1 answer:

Margaret [11]3 years ago

4 0

Answer:

OC. blizzard

Explanation:

In the United States, the National Weather Service defines a blizzard as a severe snow storm characterized by strong winds causing blowing snow that results in low visibilities. ... A severe blizzard has winds over 72 km/h (45 mph), near zero visibility, and temperatures of −12 °C (10 °F) or lower.

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What happens when a colder drier air mass pushes against a warmer moister air mass

andrew-mc [135]

A cold<span> front </span>occurs when a cold air mass advances into a region occupied by a warm air mass<span>. If the boundary between the </span>cold<span> and </span>warm air masses<span> doesn't move, it is called a stationary front. This is due to the (usually) higher humidity in the </span>air<span> of </span>warm<span> fronts compared to that of </span>cold<span> fronts.

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6 0

4 years ago

An atom of an element has 38 electrons. What is also true of that atom?

Yuliya22 [10]

A. The number of protons is the same as the number as electrons, when its just a neutral atom

6 0

4 years ago

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break

Anuta_ua [19.1K]

Answer:

$k = 9876 N/m$

Explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

$mg(h + x) = \frac{1}{2}kx^2$

also we know that maximum acceleration will be 5.3 g

so it is given as

$a = \frac{k}{m} x$

so we have

$x = \frac{ma}{k} = \frac{5.3 mg}{k}$

$mg(h + \frac{5.3 mg}{k}) = \frac{1}{2}k(\frac{5.3mg}{k})^2$

$mg(h + \frac{5.3 mg}{k}) = \frac{14.045(mg)^2}{k}$

$h + \frac{5.3mg}{k} = \frac{14.045 mg}{k}$

$h = \frac{8.745mg}{k}$

$k = \frac{8.745 (1439)(9.81)}{12.5}$

$k = 9876 N/m$

6 0

3 years ago

A 121-cm-long, 4.00 g string oscillates in its m = 3 mode with a frequency of 180 Hz and a maximum amplitude of 5.00 mm. What ar

Alex777 [14]

Answer:

0.8067 m

69.696 N

Explanation:

m = Mode = 3

M = Mass of string = 4 g

f = Frequency = 180 Hz

l = Length of string = 121 cm

Length of the string is given by

$l=m\dfrac{\lambda}{2}\\\Rightarrow \lambda=2\dfrac{l}{m}\\\Rightarrow \lambda=2\times \dfrac{1.21}{3}\\\Rightarrow \lambda=0.8067\ m$

The wavelength is 0.8067 m

Linear density is given by

$\mu=\dfrac{M}{l}\\\Rightarrow \mu=\dfrac{4\times 10^{-3}}{1.21}$

Speed of the wave

$v=f\lambda\\\Rightarrow v=180\times 2\times \dfrac{1.21}{3}\\\Rightarrow v=145.2\ m/s$

Speed of wave is given by

$v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow T=\mu v^2\\\Rightarrow T=\dfrac{4\times 10^{-3}}{1.21}\times 145.2^2\\\Rightarrow T=69.696\ N$

Tension is given by 69.696 N

8 0

3 years ago

A rocket blasts off. In 10.0 seconds it is at 10,000 ft, traveling at 3600 mph. Assuming the direction is up, calculate the acce

Andrei [34K]

Answer:

Explanation:

Givens

t = 10 seconds

vi = 0

vf = 3600 mph

a = ?

d = 10000 feet

Formula

a = (vf - vi)/t

Solution

vf = 3600 mph * 1 hr / 3600 seconds * 5280 feet / 1 mile = 5280 ft / sec

a = (5280 - 0)/10

a = 528 ft/sec^2

3 0

4 years ago