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3 years ago

6

Who studies this branch of physics? What types of careers are there in this branch of physics? What particular subjects should s

omeone study if they want a job in this branch of physics?

1 answer:

kolezko [41]3 years ago

8 0

The answers to all three questions depend on WHICH branch

of Physics you're talking about . . . a piece of information you've

neglected to specify.

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One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

2 years ago

When is a body said to be at rest

Dmitry_Shevchenko [17]

rest is the state of a body which is stationary relative to a particular frame of reference or another object. When the state of a body does not changes with time with respect to its surroundings in a particular frame of reference that time it is said to be at rest.

in other words we can say that when a body has zero velocity and its position is not changing with time relative to its surrounding then it is said to be at rest.

For example a chair on which you are sitting is at rest or a table on which your desktop is kept is rest because its position is not changing.

3 0

2 years ago

Three identical very dense masses of 3500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,

larisa86 [58]

Answer:

<em>A) 7.37 x 10^-4 N</em>

<em>B) The resultant force will be towards the -x axis</em>

Explanation:

The three masses have mass = 3500 kg

For the force of attraction between the mass at the origin and the mass -100 cm away:

distance r = 100 cm = 1 m

gravitational constant G= 6.67×10^−11 N⋅m^2/kg^2

Gravitational force of attraction $F_{g}$ = $\frac{Gm^{2} }{r^{2} }$

where G is the gravitational constant

m is the mass of each of the masses

r is the distance apart = 1 m

substituting, we have

$F_{g}$ = $\frac{6.67*10^{-11}*3500^{2} }{1^{2} }$ = 8.17 x 10^-4 N

For the force of attraction between the mass at the origin and the mass 320 cm away

distance r = 320 cm = 3.2 m

$F_{g}$ = $\frac{Gm^{2} }{r^{2} }$

substituting, we have

$F_{g}$ = $\frac{6.67*10^{-11}*3500^{2} }{3.2^{2} }$ = 7.98 x 10^-5 N

Resultant force = (8.17 x 10^-4 N) - (7.98 x 10^-5 N) = <em>7.37 x 10^-4 N</em>

<em></em>

<em>B) The resultant force will be towards the -x axis</em>

7 0

3 years ago

James and Juan were at the highest point in the football stadium, dropping water balloons on people below. Many of the balloons

DaniilM [7]

It caused the balloons to turn into physical energy

4 0

3 years ago

To withstand "g-forces" of up to 10 g's, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a "human ce

makvit [3.9K]

Answer:

31.3 m/s

Explanation:

Centripetal acceleration = v² / r where r is the length of the arm of the chair

98 m/s² = v² / r

v² = √(98 m/s² × 10) = 31.3 m/s

4 0

3 years ago