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3 years ago

As water across the surface of the earth it dissolves material as it travels. What property of water does this demonstrate?

Physics

1 answer:

iris [78.8K]3 years ago

3 0

Answer:

B) Water is a solvent

Explanation:

Among many other chemicals, water can dissolve a variety of substances. The solvent properties of chemicals are far beyond it. That property makes water a universal solvent.

Water has one atom of oxygen and two atoms of hydrogen.

One side of the water molecule is positively charged and the other side is negatively charged. This property attracts other substances like salt and disintegrates into positive and negatively charged ions. This property is due to its physical and chemical nature.

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A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

The octopus’s tentacle keeps _ right after it is bitten off ? a. Moving b. Breathing c. Growing

konstantin123 [22]

Answer:

The answer is A

Explanation:

The octopus’s tentacle keeps moving right after it is bitten off

5 0

2 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.