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blsea [12.9K]
3 years ago
9

A spacecraft is in orbit around Mars. Suppose that it was previously observed to have speed 5 km/sec while at a distance r = 500

0 km from planet center. Suppose it is now at a distance r = 6000 km from planet center. What speed would you expect it to be traveling at now? Note that for Mars we have GmMars = µMars ' 42, 828 km3 /sec2
Physics
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

V' = 4.56Km/s

Explanation:

We can use the formula of Orbital Speed, who say,

V = \sqrt{\frac{GM}{r}}

The relative velocity is given by,

V' = \sqrt{\frac{GM}{r'}}

We need to find the relation between the two speeds,

\frac{V'}{V} = \sqrt{\frac{GM}{r}*\frac{r'}{GM}}

\frac{V'}{V} = \sqrt{\frac{r}{r'}}

V' = V \sqrt{\frac{r}{r'}}

Substituting,

V' = (5)*\sqrt{\frac{5000}{6000}}

V' = 4.56Km/s

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A ray of light parallel to the principal axis of spherical mirror
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Answer:

This question is incomplete but the completed question is below

A ray of light parallel to the principal axis of a spherical mirror passes through its _____ after reflection.

(a) focus (b) pole (c) centre of curvature (d) imaginary centre of curvature

The correct option is (a)

Explanation:

After reflection, a ray of light will usually appear to diverge after they hit a spherical mirror, extending behind the mirror to meet at a point (called the focus or focal point) between the pole and the centre of curvature of this mirror.  Hence, option a is the correct option.

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3 years ago
Explain how muscles are effected by space travel
sleet_krkn [62]
Hopes this helps:

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Have a great day.
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3 years ago
What should you look for when purchasing a pair of fitness shoes?
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4 years ago
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
4 years ago
A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr
MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

7 0
3 years ago
Read 2 more answers
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