1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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>
You might be interested in
A scientist is subjected to a dose of ionizing radiation in his laboratory. without the help of radiation detection instruments,
vesna_86 [32]
<span>Health problems that develop later</span>
5 0
3 years ago
A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =
slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

7 0
3 years ago
The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob
Crazy boy [7]

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

4 0
2 years ago
If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in
Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, N_P = 50 turns

number of turns in the secondary winding, N_S = 10 turns

the secondary load resistance, R_S = 250 Ω

Determine the turns ratio;

K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5

Now, determine the reflected resistance in the primary winding;

\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0
3 years ago
Which question would most likely fill in the blank
Citrus2011 [14]
Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...
8 0
3 years ago
Other questions:
  • What is number 6 and 7 need help ASAP!!
    8·2 answers
  • Surface tension ____. a. is the inward force which tends to minimize the surface area of a liquid b. may be increased by deterge
    11·1 answer
  • While skiing in Jackson, Wyoming, your
    15·1 answer
  • The model of the universe that suggests that the sun is the center of the universe was first brought by
    5·2 answers
  • A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest
    13·1 answer
  • Given that the mass of the Earth is 5.972 * 10^24 kg and the radius of the Earth is
    8·1 answer
  • A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 458F. The stee
    13·2 answers
  • Which formula can be used to express the law of conservation of momentum, where p=momentum
    7·2 answers
  • A 10 kg bowling ball is lifted from the ground to a height of 1 meter. Assuming the bowling alley is on Earth, which type of pot
    13·1 answer
  • A sample of Iron has a mass of 46.8 grams and a volume of 6 cm 3. what is the density of iron?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!