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melomori [17]
4 years ago
5

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

, in astronomical units, AU. If planet Y is twice the mean distance from the sun as planet X, by what factor is the orbital period increased? A. 2^1/3. B. 2^1/2. C. 2^2/3. D. 2^3/2
Physics
1 answer:
kobusy [5.1K]4 years ago
3 0

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

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A scientific hypothesis can become a theory if
viktelen [127]

Answer: If, enough evidence accumulates to support a hypothesis, it moves to the next step (known as a theory) in the scientific method and becomes accepted as a valid explanation of a phenomenon.

Explanation:  a scientific theory is the framework for observations and facts.

( hope this helped )

5 0
3 years ago
During fusion, atoms are divided.<br> a. True<br> b. False
maksim [4K]
Sorry for answering late! The definition of Fusion is <span>the process or result of joining two or more things together to form a single entity. So that means its False :) </span>
8 0
3 years ago
Read 2 more answers
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
A building that is 105.50 ft. feet tall is casting a shadow that is 37.50 ft. feet. If the building next to it, is casting a sha
Arlecino [84]
115.35 ft

Set the proportion up 37.50/105.50 = 41/x and solve for x
7 0
3 years ago
Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a
Ira Lisetskai [31]

Answer:

Explanation:

Given

mass of ethanol m_e=310\ gm

mass of aluminium cup m_{al}=90\ gm

both are at an initial temperature of T_i=30^{\circ}C

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specific heat of aluminium c_{al}=0.9\ J/g-K

specific heat of ice c_i=2.108\ J/g-K

specific heat of water c_w=4.184\ J/g-K

Latent heat of fusion L=334\ J/gm

suppose m is the mass of ice added

Heat loss by Al cup and ethanol after 18^{\circ}C is reached

Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)

Heat gained by ice such that ice is melted and reached a temperature of 18^{\circ}C

Q_2=m\times 2.108\times (8.5)+m\times 334

Comparing 1 and 2 we get

m=23.65\ gm

Thus 23.65 gm of ice is added

                 

8 0
3 years ago
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