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2 years ago

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The acceleration of a particle is directly proportional to the square of the time t. When t =0, the particle is at x =24 m. Know

ing that at t =6 s, x =96 m and v =18 m/s, express x and v in terms of t.

1 answer:

AlexFokin [52]2 years ago

4 0

Answer:

x(t) = ⅟₁₀₈t⁴ + 10t + 24

v(t) = ⅟₂₇t³ + 10

Explanation:

a(t) = C₁t²

velocity is the integral of acceleration

v(t) = ⅓C₁t³ + C₂

position is the integral of velocity

x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃

x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃

C₃ = 24

x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24

72 = 108C₁ + 6C₂

C₂ = 12 - 18C₁

v(6) = 18 = ⅓C₁6³ + C₂

18 = 72C₁ + C₂

18 = 72C₁ + (12 - 18C₁)

6 = 54C₁

C₁ = 1/9

C₂ = 12 - 18(1/9)

C₂ = 10

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3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

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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 $V$ are given by:

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

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

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

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

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2 years ago

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Answer:

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Now,buyantant force

$difference \: in \: weight \: = volume(body) \times density \: of \: water \: \times g$

so;

$50 = {v}^{b} \times 1 \times {10}^{3} \times 9.8m {s}^{2}$

${v}^{b} = \frac{50}{1000 } \times 9.8$

$= \frac{50}{9800}$

$= 0.0051$

Now,

$mass \: in \: air \: = 150n = \frac{150}{9.8kg}$

$density = \frac{weght}{volume}$

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And now,

$specific \: density \: = \frac{density of \: the \: body}{density \: of \: water}$

$= \frac{3000}{1000}$

$= 3$

Hence that,specific density of a given body is 3

please mark me as brainliest, please

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2 years ago