All categories

2 years ago

Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.

Physics

1 answer:

Nookie1986 [14]2 years ago

4 0

(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.

(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

<h3>Speed of the satellite</h3>

v = √GM/r

where;

M is mass of the planet
r is radius of the planet

v = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 7,338.93 m/s

<h3>Escape velocity of the satellite</h3>

v = √2GM/r

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

<h3>Speed of the satellite at the given period </h3>

v = 2πr/T

r = vT/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.

The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

The radius of the synchronous orbit of a satellite is 69,801 km .

Learn more about minimum speed here: brainly.com/question/6504879

#SPJ1

You might be interested in

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.1 m/s. How long does h

balandron [24]

We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)

Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)

-0.31=11.1t+(1/2)(-9.81)t^2
Rearrange and solve for t
-4.905t^2+11.1t-0.31=0
t=-0.02756 or t=2.291 seconds
Reject the negative root to give
t=2.29 seconds (to 3 significant figures)

3 0

3 years ago

A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w

myrzilka [38]

The centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, $a_c = 15.625 m/s^2$ and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
$v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s$

8 0

3 years ago

if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?

Katena32 [7]

A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.

In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.

Learn more about harmonics here brainly.com/question/17315536

#SPJ10

7 0

2 years ago

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

julia-pushkina [17]

Answer:

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

$V=\pi r^2l$

$V=\pi \times 0.5^2\times 2 \ \mu m^3$

$V=1.57 \ \mu m^3$

Surface area,A

A=π d l

$A=\pi \times 1 \times 2\ \mu m^2$

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

4 0

4 years ago

How to solve arctan[tan(7pi /4)] ...?

Svet_ta [14]

Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :

<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago