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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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In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago

Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer

tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0

3 years ago

Which phrase does not describe a mineral?

Jobisdone [24]

The answer is A) specific chemical consumption

3 0

3 years ago

Read 2 more answers

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f

sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

$f=\frac{v_s}{2L}$ (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

$f=\frac{343m/s}{2(0.47228m)}=362.36Hz$

Closed tube:

$f'=\frac{v_s}{4L'}$

L': length of the closed tube = 0.702821m

$f'=\frac{343m/s}{4(0.702821m)}=122.00Hz$

Next, you use the following formula for the beat frequency:

$f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz$

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0

3 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

<h3>b)</h3>

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

<h3>c)</h3>

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

4 years ago