Bu uyguluma neden ingilizce oldu bilen varmı

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

frez [133]

Answer:

1.49 x $10^{11}$

Explanation:

Kepler's third law states that The square of the orbital period of a planet is directly proportional to the cube of its orbit.

Mathematically, this can be stated as

$T^{2}$ ∝ $R^{3}$

to remove the proportionality sign we introduce a constant

$T^{2}$ = k $R^{3}$

k = $\frac{T^{2} }{R^{3} }$

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x $10^{11}$

for mars, constant k will be

k = $\frac{687^{2} }{(2.279*10^{11}) ^{3} }$ = 3.987 x $10^{-29}$

For Earth, orbital period T is 365 days, therefore

$365^{2}$ = 3.987 x $10^{-29}$ x $R^{3}$

$R^{3}$ = 3.34 x $10^{33}$

R = 1.49 x $10^{11}$

8 0

3 years ago

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If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

$\lambda=492\ nm$

Hence, The maximum wavelength is 492 nm.

5 0

2 years ago

A car is moving with speed 60 m/s and acceleration 4 m/s2 at a given instant. (a) using a second-degree taylor polynomial, estim

pantera1 [17]

Answer:

Explanation:

Using second degree taylor polynomials

let $S(t)$ be position function and set $S(0)=0$

where S(0) is the initial position

Then $v(t) = S^i(t)$ and $a(t)=S^{ii}(t)$

we have $v(0) = 60m/s$ , $a(0) = 4m/s^2$

so $T_{2}(t) =S(0)+v(0)t+\frac{a(0)}{2} t^2\\\\\\=0+60t+\frac{4}{2} t^2\\\\T_{2}(t)=60t+2t^2\\\\S(1)=T_{2}(1)=60+2=62m$

b.) yes

6 0

3 years ago

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling

Crank

270/70^2 = x/80^2

Cross multiply

270 (6400) = 4900x

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps

4 0

2 years ago

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

levacccp [35]

Answer:

$Q = 8,345 * v$

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

$Q = 8,345 * v$

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

8 0

3 years ago

Bu uyguluma neden ingilizce oldu bilen varmı​

Bu uyguluma neden ingilizce oldu bilen varmı