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3 years ago

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Which statement best explains the relationship between the electric force between two charged objects and the distance between t

hem?
A.
As the distance increases by a factor, the electric force increases by the square of that factor.
B.
As the distance increases by a factor, the electric force increases by twice that factor.
C.
As the distance increases by a factor, the electric force decreases by twice that factor.
D.
As the distance increases by a factor, the electric force decreases by the same factor.
E.
As the distance increases by a factor, the electric force decreases by the square of that factor.

1 answer:

Thepotemich [5.8K]3 years ago

8 0

Answer: a

Explanation:

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What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not

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Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

$F=G*\frac{m1*m2}{r^{2} }$

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

$F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{2^{2} } }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{4} }$

$F'=4*G*\frac{m1*m2}{r^{2} }$

F'=4*F

<u><em> The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).</em></u>

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3 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

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Answer:

$7.1$ ms⁻¹

Explanation:

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$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

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Using conservation of angular momentum

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3 years ago

72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat

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Answer: $I=2 kg-m^2$

Explanation:

Given

mass $m=72 kg$

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Solve this physics for me <br>please with steps<br>

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Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

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F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

$[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] + [\frac{m^{2} }{s^{2} } ]$

$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

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ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the <u>capillarity</u> this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

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alekssr [168]

Answer

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