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3 years ago

8

g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron

aut wishes to age only 12 years during the round trip? (Give your answer accurate to four decimal places.)

1 answer:

Butoxors [25]3 years ago

5 0

Answer:

The value is $v = 2.999 *10^{8} \ m/s$

Explanation:

From the question we are told that

The time taken to travel to the planet from earth is $t = 189 \ light-years$

The time to be spent on the ship is $t_{s} = 12 \ years$

Generally speed can be obtained using the mathematical relation represented below

$t_s = 2 * t * \sqrt{1 - \frac{v^2}{c^2 } }$

The 2 in the equation show that the trip is a round trip i.e going and coming back

=> $12 = 2 * 189 * \sqrt{1 - \frac{v^2}{(3.0*10^{8})^2 } }$

=> $v = 2.999 *10^{8} \ m/s$

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The physical law that explains that is the law of conservation of energy which states that he energy of an isolated sistem remains constant

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3 years ago

A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

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3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

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Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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3 years ago

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kkurt [141]

Answer:

Explanation:

for baseball

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Where, I be the moment of inertia and ω be the angular speed.

ω = v/r

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(b) Let the velocity of juice can is v'.

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3 years ago

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of

Lana71 [14]

Electric field due to a point charge is given as

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also the distance is given as

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now we will have

$E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}$

so we will have

$E = 5.14 \times 10^{11} N/C$

so above is the electric field due to proton

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3 years ago