Physics, 01.10.2019 00:10 AM28

Six artificial satellites circle a space station at constant speed. the mass m of each satellite, distance l from the space station, and the speed v of each satellite are listed below. the satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. the gravitational force is negligible. 1. m=200kg, l= 5000 m, v=120 m/s 2. m=800kg, l= 10,000 m, v=40m/s 3. m=400kg, l= 2500 m, v=80m/s 4. m=100kg, l=2500 m, v=160m/s 5. m=300kg, l=10,000m, v=80 m/s 6. m=200kg, l=5000 m, v=160 m/s part a rank each satellite from largest to smallest based on its period. part b rank each satellite from largest to smallest based on its acceleration.

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Answer from: shan8793

A) Ranking of each satellite from largest to smallest based on its period is;

T2 > T5 > T1 > T3 = T6 > T4

B) Ranking of each satellite from largest to smallest based on its acceleration is;

a4 > a6 > a1 > a3 > a5 > a2

A) We want to rank them based on their period;

Formula for period here is;

T = 2πL/v

Thus;

Satellite 1; m = 200kg,L= 5000 m, v = 120 m/s;

T1 = 2π × 5000/120

T1 = 261.8 s

Satellite 2; m = 800kg, L = 10,000 m, v=40m/s;

T2 = 2π × 10000/40

T2 = 1570.8 s

Satellite 3; m = 400kg, L = 2500 m, v = 80m/s ;

T3 = 2π × 2500/80

T3 = 196.35 s

Satellite 4; m = 100kg, L = 2500 m, V = 160m/s

T4 = 2π × 2500/160

T4 = 98.175 s

Satellite 5; m = 300kg, L = 10,000m, V = 80 m/s

T5 = 2π × 10000/80

T5 = 785.4 s

Satellite 6; m = 200kg, L = 5000 m, V = 160 m/s

T6 = 2π × 5000/160

T6 = 196.35 s

Ranking from largest to smallest Period is;

T2 > T5 > T1 > T3 = T6 > T4

B) We want to rank them based on their period;

Formula for acceleration here is;

a = v²/L

Thus;

Satellite 1; a1 = 120²/5000

a1 = 2.88 m/s²

Satellite 2; a2 = 40²/10000

a2 = 0.16 m/s²

Satellite 3; a3 = 80²/2500

a3 = 2.56 m/s²

Satellite 4; a4 = 160²/2500

a4 = 10.24 m/s²

Satellite 5; a5 = 80²/10000

a5 = 0.64 m/s²

Satellite 6; a6 = 160²/5000

a6 = 5.12 m/s²

Ranking from largest to smallest gives;

a4 > a6 > a1 > a3 > a5 > a2

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Answer from: jnaifuson

Answers:

a) $T_{2}T_{5}T_{1}T_{3}=T_{6}T_{4}$

b) $a_{4}a_{6}a_{1}a_{3}a_{5}a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

is the radius of the orbit of each satellite

is the period of the orbit of each satellite

Isolating :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}T_{5}T_{1}T_{3}=T_{6}T_{4}$

b) Acceleration is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}a_{6}a_{1}a_{3}a_{5}a_{2}$

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.