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Makovka662 [10]
3 years ago
14

In developed nations, the water delivered to homes by pipe is generally potable. Please select the best answer from the choices

provided T F
Physics
2 answers:
andrey2020 [161]3 years ago
8 0

Answer: The given statement is true.

Explanation:

In developed nations, water delivered to homes is purified and passes through different stages to purification so that people will get clean and safe water.

This also prevents any type of diseases cause by drinking dirty or non-purified water. For example, cholera, typhoid, dysentery etc are caused by drinking dirty water.

Thus, we can conclude that the statement in developed nations, the water delivered to homes by pipe is generally potable, is true.

stiv31 [10]3 years ago
6 0

True is ur answer sorry if i am too late

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17.Explain the different ways that an object can become electrically charged.
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17.

There are three different methods for charging objects:

- Friction: in friction, two objects are rubbed against each other. As a result, electrons can be passed from one object to the other, so one object will gain a net negative charge while the other object will gain a net positive charge due to the lack of electrons.

- Conduction: this occurs when two conductive objects are put in contact with each other, and charges (electrons, usually) are transferred from one object to the other one.

- Induction: this occurs when two objects are brought closer to each other, but not in contact. If one of the two objects has a net charge (different from zero) on its surface, then it will induce a movement of charges in the second object: in particular, in the second object, charges of the opposite polarity will be attracted towards the first object, while charges of same polarity will be repelled further away.

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Charged objects produce around themselves an electric field. The strenght of the electric field is given by (assuming the charged objects are spherical)

E=k\frac{q}{r^2}

where k is the Coulomb's constant, q is the magnitude of the charge and r the distance from the centre of the charge. As we see, the strength of the field is inversely proportional to the square of the distance.

Also, the direction of the field is determined by the sign of the charge:

- if the charge is positive, the electric field points away from the charge (this means that other positive charges in the field will be repelled away)

- if the charge is negative, the electric field points towards the charge (this means that other positive charges in the field will be attracted towards it)

19.

Electrical force is given by:

F=k\frac{q_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 are the two charges, and r their separation.

Gravitational force is given by:

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r their separation.

Similarities between the two forces:

- Both are inversely proportional to the square of the distance between the two objects, r

- Both are non-contact forces (the two objects can experience the forces even if they are not in contact)

- Both forces have infinite range

Differencies between the two forces:

- The electric force can be either attractive or repulsive, while the gravitational force is attractive only

- The electric force is much stronger than the gravitational force, due to the much larger value of the Coulomb's constant k compared to the gravitational constant G

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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

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