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2 years ago

Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.

Physics

1 answer:

Gemiola [76]2 years ago

7 0

Answer:

a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

V = i R

i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

R = 4 Ω

i₄ = V / 4

R = 8 Ω

i₈ = V / 8

we look for the relationship between these two currents

i₈ /i₄ = 4/8 = ½

i₈ = 0.5 i₄

R = 3 Ω

i₃ = V3

R = 10 Ω

i₁₀ = V / 10

we look for relationships

i₁₀ / 1₃ = 3/10

i₁₀ = 0.3 i₃

i₁₀ / 1₈ = 8/10

i₁₀ = 0.8 i₈

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A ball dropped from a height of 50 meters, Determine the speed of the ball after 3 seconds

Alex17521 [72]

16.7 m/s

To determine the speed we use the formula: Distance ÷ Time so, 50 ÷ 3 = 16.66... ( rounded off to 1.d.p is 16.7).

8 0

3 years ago

A. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.

never [62]

Answer:

a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

F₁ > F₃

therefore the correct answer is 3

3 0

3 years ago

Two identical stars with mass M orbit around their center of mass. Each orbit is circular and has radius

kondaur [170]

Answers: (A) $F=G\frac{M^2}{4R^2}$ (B) $V=\sqrt{\frac{GM}{4R}}$ (C) $T=4\pi R\sqrt{\frac{R}{GM}}$ (D)

$E=-\frac{GM^{2}}{4R}$

Explanation:

<h2>(A) Gravitational force of one star on the other</h2>

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the gravitational force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In the case of this binary system with two stars with the same mass $M$ and separated each other by a distance $2R$ , the gravitational force is:

$F=G\frac{(M)(M)}{(2R)^2}$ (2)

$F=G\frac{M^2}{4R^2}$ (3) This is the gravitational force between the two stars.

<h2>(B) Orbital speed of each star</h2>

Taking into account both stars describe a circular orbit and the fact this is a symmetrical system, the orbital speed $V$ of each star is the same. In addition, if we assume this system is in equilibrium, gravitational force must be equal to the centripetal force $F_{C}$ (remembering we are talking about a circular orbit):

So: $F=F_{C}$ (4)

Where $F_{C}=Ma_{C}$ (5) Being $a_{C}$ the centripetal acceleration

On the other hand, we know there is a relation between $a_{C}$ and the velocity $V$ :

$a_{C}=\frac{V^{2}}{R}$ (6)

Substituting (6) in (5):

$F_{C}=M\frac{V^{2}}{R}$ (7)

Substituting (3) and (7) in (4):

$G\frac{M^2}{4R^2}=M\frac{V^{2}}{R}$ (8)

Finding $V$ :

$V=\sqrt{\frac{GM}{4R}}$ (9) This is the orbital speed of each star

<h2>(C) Period of the orbit of each star</h2><h2 />

The period $T$ of each star is given by:

$T=\frac{2\pi R}{V}$ (10)

Substituting (9) in (10):

$T=\frac{2\pi R}{\sqrt{\frac{GM}{4R}}}$ (11)

Solving and simplifying:

$T=4\pi R\sqrt{\frac{R}{GM}}$ (12) This is the orbital period of each star.

<h2>(D) Energy required to separate the two stars to infinity</h2>

The gravitational potential energy $U_{g}$ is given by:

$U_{g}=-\frac{Gm_{1}m_{2}}{r}$ (13)

Taking into account this energy is always negative, which means the maximum value it can take is 0 (this happens when the masses are infinitely far away); the variation in the potential energy $\Delta U_{g}$ for this case is:

$\Delta U_{g}=U-U_{\infty}$ (14)

Knowing $U_{\infty}=0$ the total potential energy is $U$ and in the case of this binary system is:

$U=-\frac{G(M)(M)}{2R}=-\frac{GM^{2}}{2R}$ (15)

Now, we already have the potential energy, but we need to know the kinetic energy $K$ in order to obtain the total Mechanical Energy $E$ required to separate the two stars to infinity.

In this sense:

$E=U+K$ (16)

Where the kinetic energy of both stars is:

$K=\frac{1}{2}MV^{2}+\frac{1}{2}MV^{2}=MV^{2}$ (17)

Substituting the value of $V$ found in (9):

$K=M(\sqrt{\frac{GM}{4R}})^{2}$ (17)

$K=\frac{1}{4}\frac{GM^{2}}{R}$ (18)

Substituting (15) and (18) in (16):

$E=-\frac{GM^{2}}{2R}+\frac{1}{4}\frac{GM^{2}}{R}$ (19)

$E=-\frac{GM^{2}}{4R}$ (20) This is the energy required to separate the two stars to infinity.

4 0

2 years ago

Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb

melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

$F=G\frac{Mm}{r^2}$

$F=G\frac{M6m}{r_b^2}$

$G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r$