Ask question

Ask question

All categories

3 years ago

6

Severe weather often includes destructive events such as Hurricanes and Tornadoes where air is moving much faster than usual. Tr

ue or False

2 answers:

xeze [42]3 years ago

7 0

Answer:

True

Explanation:

I just took the quiz and got it right.

lapo4ka [179]3 years ago

7 0

The correct answer is True

Explanation:

The term "severe weather" is mainly used to describe extreme weather conditions or phenomena that are dangerous or destructive. In this context, both hurricanes and tornadoes are part of severe weather because in both, air spins abnormally and there are abnormal precipitations that can be destructive for humans and ecosystems. Besides this, other such as thunderstorms, floods or blizzards are part of severe weather. According to this, the statement presented is true.

You might be interested in

How is temperature related to the physical change of a<br>substance

blagie [28]

Answer:

~~Now, you have left your question very open ended and didn't ask for any particular kind of answer so I'll do my best to get what you're looking for.~~

A physical change in a substance doesn't change what the substance is. It can possibly melt or freeze an object. I mean heat makes things expand while cooling makes them retract.... In chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

4 0

3 years ago

Read 2 more answers

What is the jet stream and how does it separate cold air to warm air?

docker41 [41]

Jet stream<span>ˈjet ˈˌstrēm/</span>noun1.a narrow, variable band of very strong, predominantly westerly air currents encircling the globe several miles above the earth. There are typically two or three jet streams in each of the northern and southern hemispheres.
(credit to google)

4 0

4 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

belka [17]

Given that,

Energy $H=2.7\times10^{31}\ W$

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

$H=Ae\sigma T^4$

$A=\dfrac{H}{e\sigma T^4}$

$4\pi R^2=\dfrac{H}{e\sigma T^4}$

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}$

$R=5.0\times10^{10}\ m$

(b). Given that,

Radiates energy $H=2.1\times10^{23}\ W$

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}$

$R=5.42\times10^{6}\ m$

Hence, (a). The radius of the star is $5.0\times10^{10}\ m$

(b). The radius of the star is $5.42\times10^{6}\ m$

8 0

3 years ago

A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is

OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

$U = \frac{GMm}{r}$

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

$U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}$

$U_2= - 5.1180*10^9J$

Point B )

We now use the average radius distance from the earth.

$U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}$

$U_2= -5.8925*10^9J$

Then,

$\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)$

$\Delta U = 774.5*10^6$

By the law of conservation of energy we know that,

$\Delta U = \frac{1}{2}mv^2$

clearing v,

$v= \sqrt{2 \Delta U/m}$

$v= \sqrt{2*774.5*10^6 /94.2}$

$v= 4055.08m/s$

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0

3 years ago

As the temperature of a liquid increases, the solubility of a solid in the liquid

nexus9112 [7]

D, it depends. For water, the solubility increases as it get hotter, up to 100 degrees, but in some cases, liquids behave the opposite way.

4 0

3 years ago