Ask question

Ask question

All categories

3 years ago

15

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10

,000 kg, and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length 3.0 m each and mass 10 kg. The antenna’s lie in the plane of rotation. What is the angular momentum of the satellite?

1 answer:

bearhunter [10]3 years ago

5 0

Answer:

605447.7066 kgm²/s

Explanation:

$m_1$ = Mass of sphere = 10000 kg

$m_2$ = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

$\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s$

Angular momentum is given by

$L=I\omega$

Moment of inertia of the satellite is

$I_s=\frac{2}{5}m_1r^2$

Moment of antenna of the satellite is

$I_a=\frac{1}{3}m_2l^2$

The angular momentum of the system is

$L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s$

The angular momentum of the satellite is 605447.7066 kgm²/s

You might be interested in

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

asambeis [7]

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$ , $m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$ , $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$ .

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

3 0

3 years ago

Dos bolas, de masas mA = 40 g y mB =60 g, esta?n suspendidas como se observa en la figura. La bola ma?s ligera se jala en un a?n

Ymorist [56]

MA=40 or something I really don’t know

3 0

3 years ago

You are In-line skates at the top of a small hill. Your potential energy is equal to 1000 J. The last time we checked, your mass

stira [4]

A) <u>Weight = mass × acceleration (due to gravity) </u>

= 60×9.8

= 588 N

<u>B) Potential energy = mass x gravity x change in height </u>

1,000 = 60.0 x 9.8 x h

h = 1.7 m

<u>C) Kinetic energyF = potential energyI </u>

KEF = 1/2mv2

PEI = mgh = 1,000 J

1/2mv2 = 1,000

1/2(60.0)v2 = 1,000

v2 = 33.33

v = 5.77 m/s

3 0

4 years ago

Hey guys can you help me solve this problem "how long will it take a car travelling 30m/s to come to stop ifs its acceleration i

salantis [7]

Answer:

10 seconds.

Explanation:

We can use a kinematic equation where we know the final velocity, initial velocity, acceleration, and need to determine the time <em>t: </em>

<em /> $\displaystyle v_f = v_i + at$ <em />

<em />

The initial velocit is 30 m/s, the final velocity is 0 m/s (as we stopped), and the acceleration is -3 m/s².

Substitute and solve for <em>t: </em>

<em /> $\displaystyle \begin{aligned} (0\text{ m/s}) & = (30 \text{ m/s}) + (-3 \text{ m/s$^2$}) t \\ \\ t & = \frac{-30\text{ m/s}}{-3 \text{ m/s$^2$}} \\ \\ & = 10 \text{ s} \end{aligned}$ <em />

<em />

Hence, it will take the car 10 seconds to come to a stop.

7 0

2 years ago

two ice skaters push each other, and move in opposite directions. the mass of one is 75 kg, and his speed is 2 m/s. if the mass

sattari [20]

momentum conservation

75x2 = 30 x her speed

75x2/30=150/30=5m/s

7 0

4 years ago