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3 years ago

What is the length of the X component of the vector plotted below

Physics

1 answer:

dlinn [17]3 years ago

3 0

Answer:

Explanation:

If you straighten out the line, it touches the 5, which makes the length 5

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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

Explanation:

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

3 years ago

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

siniylev [52]

Answer:

The height of the object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of $Gmm_{e}$

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$200=\dfrac{Gmm_{e}}{(3958.756)^2}$

$200\times(3958.756)^2=Gmm_{e}$

$Gmm_{e}=3.134\times10^{9}$

We need to calculate the height of the object

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$125=\dfrac{200\times(3958.756)^2}{r^2}$

$r^2=\dfrac{200\times(3958.756)^2}{125}$

$r^2=25074798.5$

$r=\sqrt{25074798.5}$

$r=5007.4\ miles$

Hence. The height of the object is 5007.4 miles.

7 0

3 years ago

BEST ANSWER GETS BRAINLIEST! WORTH LOTS OF POINTS!!

Lelechka [254]

The energy is being converted to the water yet it isn't rising because the water is absorbing the heat not reflecting it. I think this is correct but I'm not sure. I would like to know the answer if you find out though.

6 0

3 years ago

Read 2 more answers

Someone help me?

kap26 [50]

Answer:

-27.3 m/s

Explanation:

Given:

y₀ = 38 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (0 m − 38 m)

v = -27.3 m/s

Or, you can solve with energy.

PE = KE

mgh = ½ mv²

v² = 2gh

v = -27.3 m/s

3 0

3 years ago

What minimum speed must an electron have to excite the 492-nm-wavelength blue emission line in the hg spectrum?

mezya [45]

The energy required by the excitation of the line is:
ΔE = hν = hc / λ
where:
ΔE = energy difference
h = Planck constant
ν = line frequency
c = speed of light
λ = line wavelength

The energy difference must be supplied by the electron, supposing it transfers all its kinetic energy to excite the line:
$\Delta E = \frac{1}{2} m v^{2}$

Therefore,
$\frac{1}{2} m v^{2} = \frac{hc}{\lambda}$

And solving for v we get:
$v = \sqrt{ \frac{2hc}{m\lambda} }$

Plugging in numbers (after trasforing into the correct SI units of measurement):
$v = \sqrt{ \frac{(2)(6.6 \cdot 10^{-34})(3 \cdot 10^{8}) }{(9.11 \cdot 10^{-31})(4.92 \cdot 10^{-7}) } }$
=9.4 · 10⁵ m/s

Hence, the electron must have a speed of 9.4 · 10⁵ m/s in order to excite the 492nm line.

5 0

3 years ago

What is the length of the X component of the vector plotted below￼

What is the length of the X component of the vector plotted below