D

Most stars are<br> a.brown fs<br> b.main sequence<br> c.white dwarfs<br> d.pulsars

trapecia [35]

I believe the answer is B. Main sequence

3 0

3 years ago

when two resistors are wired in series with a 12 v battery, the current through the battery is 0.31 a. when they are wired in pa

liberstina [14]

Let R₁ and R₂ be the two Resistance

R₁ - Resistance of 1st resistor

R₂ - Resistance of 2nd resistor

V = Voltage = 12V

I = O.31A ( in series)

I = 1.6 A ( in parallel)

when two resistance connected in series

Rs = R₁ + R₂

V = IRs = 12 /0.31 V

R₁+R₂ = 38.700Ω (equation1.)

When two resistance connected in parallel

$R_{p} =\frac { R_{1} R_{2}}{R_{1} + R_{2} }$

V = I Rp

V= I (R₂ R₁ /R₁ +R₂)

R₁ R₂ = 290.25 Ω

As we know

(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2

Using above values we get,

(R₁ - R₂ ) ² = (38.70) ² - 4x 290.25

(R₁ - R₂ ) ² = 336.69 Ω

R₁ - R₂ = 18.34 (equation 2 )

Using equation 1 and 2 we get

R₁+ R₂ = 38.70 52

R₁ - R₂ = 18.34

R₁ = 28.535 Ω

Using the value of R₁ in equation 1 we get

R₂ = 38·700 - 28.535

RR₂ = 10.125 Ω

R₁ = 28. 535 v and R₂ 10.125 Ω

To know more about resistance in series and parallel:

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7 0

1 year ago

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?

SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0

3 years ago

A fence 8 ft high (w) runs parallel to a tall building and is 24 ft (d) from it. Find the length (L) of the shortest ladder t

AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

$Hypotenuse^2 = opposite^2 + adjacent^2$

The hypotenuse is the length of the ladder.

$Hypotenuse = \sqrt{opposite^2 + adjacent^2}$

$L^2 = BC^2 + (24 + AE)^2$ ........1

Triangle ABC is similar to triangle AEF, hence:

$\frac{BC}{8} = \frac{AE + 24}{AE}$

BC = $\frac{8(AE + 24)}{AE}$ .................2

Substitute 2 into 1

$L^2$ = $(\frac{8(AE + 24)}{AE})^2$ + $(24 + AE)^2$

Let AE = x

$L^2$ = $(\frac{8x + 192}{x})^2 + (24 + x)^2$

= $(8 + \frac{192}{x})^2 + (24 + x)^2$

Minimize L with respect to x.

2 $\frac{dL}{dx}$ = $2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)$

=

5 0

3 years ago

Two equal quantities of water, of mass m and at temperatures T1 and T2, (T1 > T2) are mixed together with the pressure kept c

Ray Of Light [21]

Answer:

ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))

Explanation:

The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.

4 0

3 years ago