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sertanlavr [38]
3 years ago
9

NASA decides to put a 128 kg satellite into orbit over the planet Venus because they want to take pictures. The satellite is 37,

000,000 m above the surface of venus. 
Calculate the gravitational field at that altitude. m/s2
how do you calculate this
Physics
1 answer:
ycow [4]3 years ago
6 0
1).  Look up the acceleration of gravity on the surface of Venus.  It's 8.87 m/s².
(That's about 10% less than on Earth.)

2).  Look up the radius of Venus.  It's 6051.8 km.  
(That's about 5% less than Earth's.)

3).  Remember that the gravitational field (acceleration, force) changes
opposite to the square of the distance from the planet ... or the distance
between any two masses that are gravitating.

Now we have enough information to do the calculating.  Notice that the question
only asks for the planet's "gravitational field" (acceleration) way out there.  That
has nothing to do with the satellite's mass, or whatever you decide to put out there,
or even if there's nothing out there at all.  It's just a characteristic of Venus at that
distance from it.

The distances we need to compare are the distances from the center of Venus.

On the surface (distance from the center is the radius of Venus), it's 6,051,800 m.
In the orbit, it's 37,000,000 m.

The ratio is (37,000,000 / 6,051,800) = about 6.1 .

The gravity way out in the orbit is less than on the surface, and by the
same amount as the square of the distance ratio.

Surface gravity = 8.87

Gravity out at the orbit = 8.87 divided by (the distance ratio)²

Gravity = 8.87 / (6.1)²

Better way:  Gravity = 8.87 x (6,051,800/37,000,000)² = <em>0.2373 m/sec²</em>
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In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla
Harlamova29_29 [7]

Answer:

\lambda'=78.086\ nm

Explanation:

Given:

  • wavelength of light in the air, \lambda=530\times 10^{-9}\ m
  • time taken to travel from the source to the photocell via air, t=16.7\ s
  • time taken to reach the photocell via air and glass slab, t'=21.3\times 10^{-9}\ s
  • thickness of the glass slab, x=0.87\ m

<u>Now we have the relation for time:</u>

\rm time=\frac{distance}{speed}

hence,

t=\frac{d}{c}

c= speed of light in air

16.7\times 10^{-9}=\frac{d}{3\times 10^8}

d=16.7\times 10^{-9}\times 3\times 10^8

d=5.01\ m

For the case when glass slab is inserted between the path of light:

\frac{(d-x)}{c} +\frac{x}{v} =t' (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}

v=4.42\times 10^7\ m.s^{-1}

Using Snell's law:

\frac{\lambda}{\lambda'} =\frac{c}{v}

\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}

\lambda'=78.086\ nm

5 0
3 years ago
What is the best definition of luminous?
vagabundo [1.1K]

Answer:

the state of giving off light or glow.

8 0
3 years ago
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People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p
Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

g' = \omega^2 (\dfrac{r}{2})

g' =\dfrac{1}{2}(\omega^2 r)

g'=\dfrac{g}{2}

People at the halfway experience g/2

hence, the correct answer is option B

7 0
3 years ago
Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
aliina [53]

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

7 0
3 years ago
8
Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0
3 years ago
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