All categories

3 years ago

HURRYYYWhich tools do meteorologists use to collect data about the weather?

Physics

2 answers:

Andrei [34K]3 years ago

4 0

Weather satellites are used to get data about weather

lys-0071 [83]3 years ago

3 0

Answer:

satellites and weather fronts

You might be interested in

Fusion occurs with atoms of elements less massive than

Nimfa-mama [501]

The answer is A. Iron.

4 0

3 years ago

Which of the following is NOT NECESSARILY a property of an air mass?

lilavasa [31]

Answer:

Explanation:

The property of air mass include

1) it must be large.

2) it must have relatively uniform properties.

3)it must travel as a recognizable entity.

It must have a warm front at its leading edge, is not necessarily a property of an air mass because Not all air masses have a warm front at their leading edge. There are five types of air masses and different types of the front can be developed.

4 0

4 years ago

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

b) Density at 22 degrees: $1.355\cdot 10^4 kg/m^3$

We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$

8 0

3 years ago

Which symbol is used for an alpha particle? a. β c. γ b. Δ d. α

xenn [34]

It is the very last answer

6 0

3 years ago

How much force is needed to accelerate a 70 kg rider and her 200 kg

strojnjashka [21]

Answer: f= mass×acceleration

They said t1he scooter so 200× 4

=800N

Explanation:

I am not that sure

7 0

3 years ago