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3 years ago

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A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from

the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:

1 answer:

mamaluj [8]3 years ago

7 0

Answer:

The expected year is 2017.

Explanation:

Total years that the millionaire to live = 15 years

Travel away from the earth at = 0.8 c

This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:

$T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years$

Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017

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A car, initially traveling 28.0ft/s, steadily speeds up to 50.0ft/s in 7.40s. Determine all unknowns and answer the following qu

nataly862011 [7]

Explanation:

Given:

$v_0 = 28.0\:\text{ft/s}$

$v = 50.0\:\text{ft/s}$

$t = 7.40\:\text{s}$

First, we calculate the acceleration of the car during this time:

$v = v_0 + at \Rightarrow a = \dfrac{v - v_0}{t}$

Plugging in the given values, we get

$a = \dfrac{50.0\:\text{ft/s} - 28.0\:\text{ft/s}}{7.40\:\text{s}} = 2.97\:\text{ft/s}^2$

Now that we have the value for the acceleration, we can solve for the distance traveled during the time t:

$x = v_0t + \frac{1}{2}at^2$

$\:\:\:\:=(28.0\:\text{ft/s})(7.40\:\text{s})$

$\:\:\:\:\:\:\:\:\:\:\:\:+ \frac{1}{2}(2.97\:\text{ft/s}^2)(7.40\:\text{s})^2$

$\:\:\:\:= 289\:\text{ft}$

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A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista

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Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

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v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

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A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

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