All categories

3 years ago

What occurs in an updraft

2 answers:

5 0

Updrafts and downdrafts alsooccur as part of the turbulence that is created when air passes over topographic barriers such as mountains. ... Updraftscharacterize a storm's early development, during which warm air rises to the level where condensation begins and precipitation starts to develop.

mezya [45]3 years ago

4 0

Correct answer: Warm air from the south rises.

You might be interested in

Which of the following codes is used to report the removal of 37 skin tags by electrosurgical destruction?

Tema [17]

Are there any answer choices??

7 0

3 years ago

Vectors are an important part of the language of science, mathematics, and engineering.

tensa zangetsu [6.8K]

Answer:

a. True

Explanation:

Vectors are an important part of the language of science, mathematics, and engineering.

4 0

2 years ago

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago

Two charges repel each other with a force. If the magnitude of both charges double, but the distance between them remains consta

-Dominant- [34]

Answer:

Explanation:

4 0

3 years ago