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Lorico [155]
3 years ago
10

Find equivalent resistance. Answer asap and please, please don't spam.​

Physics
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega

So, the equivalent resistance s 4.77 ohms.

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Answer:

Explanation:

  a )

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putting the values

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b )

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lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

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c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

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5 0
3 years ago
For a given amount of gas at a constant temperature, the volume of a gas varies inversely with its pressure is a statement of __
Ksivusya [100]

Answer:

d. Boyle's

Explanation:

Boyle's Law: States that the volume of a fixed mass of gas is inversely proportional proportional to its pressure, provided temperature remains constant.

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V = k/P, Where k is the constant of proportionality

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7 0
3 years ago
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different
kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.

Solution:

- The standard atmospheric equation is expressed as:

                           P = P_a* ( 1 - βh/T_a)^(g / R*β)

                          (g / R*β) = 32.174 / 1716*0.0035 = 5.252

                            P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

                            P = 1240 lb/ft^2

- The air density method which is expressed as:

                            P = P_a - γ*h

                            P = 2116.2 - 0.07647*14,110

                            P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

                            P = P_a* e^ ( -g*h / R*T_a )

                            P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

                            P = 1270 lb/ft^2

6 0
3 years ago
Read 2 more answers
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