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2 years ago

10

The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that

have charges or partial charges are __________.

1 answer:

Arlecino [84]2 years ago

4 0

Answer:

The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are <u>due to water's partial charges.</u>

Explanation:

The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.

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_____ are placed on dangerous machinery to detect motion, light, heat, pressure, or another stimulus. Their presence helps prote

goblinko [34]

Answer;

-Sensors

-Sensors are placed on dangerous machinery to detect motion, light, heat, pressure, or another stimulus. Their presence helps protect operators from injury while working on machines.

Explanation;

-Machinery, safety and factory floor sensors and switches help workers become more productive, efficient, and safe.

-Hazardous machines and systems are frequently equipped with safety elements (safety doors) with a locking mechanism to protect the operator. Their function is to prevent hazardous machine functions if the safety door is not closed and locked and to keep the safety door closed and locked until the risk of injury has passed.

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2 years ago

Read 2 more answers

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

2 years ago

With a bit of algebraic reasoning find your gravitational acceleration toward any planet of mass M a distance d from its center.

grandymaker [24]

The acceleration due to gravity is given as:

g = GM/r²

<h3>Derivation of gravitational acceleration:</h3>

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

F<em>g </em>= GMm/(r + h)²

F<em>g = </em>gravitational force

From Newton's second law of motion,

F<em>g </em>= ma

a = F<em>g</em>/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

Learn more about gravitational acceleration here:

brainly.com/question/2142879

#SPJ4

5 0

1 year ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

So

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

7 0

2 years ago

A research submarine has a 20-cm-diameter window that is 9.0 cm thick. The manufacturer says the window can withstand forces up

Ratling [72]

Answer:

The maximum safe depth in salt water is 3758.2 m.

Explanation:

Given that,

Diameter = 20 cm

Radius = 10 cm

Thickness = 9.0 cm

Force $F = 1.2\times10^{6}\ N$

Inside pressure = 1.0 atm

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

Put the value into the formula

$A=\pi\times(10\times10^{-2})^2$

$A=0.0314\ m^2$

We need to calculate the pressure

Using formula of pressure

$P=\dfrac{F}{A}$

Put the value into the formula

$P=\dfrac{1.2\times10^{6}}{0.0314}$

$P=38216560.50\ Pa$

$P=3.8\times10^{7}\ Pa$

We need to calculate the maximum depth

Using equation of pressure

$P=P_{atm}+\rho gh$

$h=\dfrac{P-P_{atm}}{\rho g}$

Put the value into the formula

$h=\dfrac{3.8\times10^{7}-101325}{1029\times9.8}$

$h=3758.2\ m$

Hence, The maximum safe depth in salt water is 3758.2 m.

4 0

2 years ago