Please need help on this

Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo

Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c) d) w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

T = 1 / f

f = 1 / T

f = 1 / 3.6 104

f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

d = 2pi (R + r)

d = 2pi (69 911 103 + 120 106)

d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

.w = 2pi /t

w = 2pi / 3.6 10⁴

w = 1.745 10⁻⁴ rad / s

how fast is

v = w r

v = 1.75 10-4 (69.911 106 + 120 106)

v = 190017 m / s

5 0

3 years ago

A circular horizontal surface having an area of 1020.5 mm2that is completely immersed in a fluid experiences a uniform force of

Bingel [31]

Answer:

Faya vei Abhikesh brass hahahahahaha

Explanation:

3 0

3 years ago

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis

Elza [17]

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1 ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

4 0

3 years ago

How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water

alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0

3 years ago

A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se

Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

$x = 32t - 38\frac{t^3}{3} \\\\$

The velocity of the particle is calculated as the change in the position of the particle with time;

$v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s$

Acceleration is the change in velocity with time;

$a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2$

4 0

2 years ago