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2 years ago

Which "spheres" are interacting when water evaporates from plants

2 answers:

8 0

The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.

Volgvan2 years ago

4 0

Answer:

Explanation:

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Please Help Me!!!

vova2212 [387]

Answer:

There are the molecular changes water undergoes

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3 years ago

A group of scientists decide to repeat the muon decay experiment (TR Section 2.7) at the Mauna Kea telescope site in Hawaii, whi

m_a_m_a [10]

Answer:They decided to count 10000 muons because of classical predictions. When you count 1000 initial muons.they would expect only 1.5 muons

Explanation:

t= 4205/0.98 = 1.43×10^-5sec

Classically,number of surviving muons= No exp(-0.693t/t)

N=10^4exp[(-0.694×(1.43×10^-5)/1.52×10-6)]

N= 14.7=15

Relativistic time t'=t/y

t'=(1.43×10^-5)×sqrt(1-(0.98c)^2/c^2

t'=2.84×10^-6secs

N= 10^4exp[(-0.693×(3.84×10^-6)/1.52×10^-6]

N relativistic=2739

4 0

3 years ago

3. A -4.00-uC charge lies 20.0cm to the right of a 2.00-uC charge on the x axis. What is the force on the 2.00-uC charge?

Sladkaya [172]

<u>Answer:</u>

The force on the 2.00-uC charge is $36 \times 10^{10} \mathrm{N}$

<u>Explanation:</u>

We know that force between two charges is given by Coulomb’s law,

$\mathrm{F}=\mathrm{k} \frac{q 1 q 2}{r * r}$

Where k = Coulomb’s constant =

$9.0 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$

And q1 and q2 are the charges given to be = -4.00-uC and 2.00-uC charges

And r = distance between the charges = 20 cm = 0.2 m

Substituting the given values in the formula we get force applied on $2.00\ \mu C$ charge,

F = $36 \times 10^{10} \mathrm{N}$ attractive force which is the required answer.

3 0

2 years ago

What is an inexpensive, portable, and common way to assess body fat in the fitness industry?

Musya8 [376]

Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

5 0

2 years ago

A charge of +2.00 x 10^-9 C is placed at the origin, and another charge of +4.50 x 10^-9 C is placed at x = 1.6 m. The Coulomb c

igor_vitrenko [27]

Answer:

0.64 m from the first charge

Explanation:

Force is given by

$F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}$

$F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}$

These forces are equal

$\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m$

The distance that charge should be placed is 0.64 m from the first charge

3 0

3 years ago