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Trava [24]
3 years ago
6

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

ng the cavity to force out the water through an opening. A 6.50 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. A. If the squid has 1.55 kg of water in its cavity, at what speed must it expel this water to suddenly achieve a speed of 2.60 m/s to escape the predator? Neglect any drag effects of the surrounding water.
B. How much kinetic energy does the squid create by this maneuver?
Physics
1 answer:
Alex3 years ago
4 0

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

Momentum gained by squid = P=m_1v_1

Mass of water = m_2=1.55 kg

Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

m_1v_1=m_2v_2

v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}

=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules

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Explanation:

We know that that the range of the ball on the earth

R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}

therefore, range of the ball on moon

R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}

R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}

therefore,

R_{moon}=6R_{earth}

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2 years ago
On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo
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Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

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How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances
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Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

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M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

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A 100 W lightbulb produces 5.0 W of visible light. (The other 95 W are dissipated as heat and infrared radiation.) What is the l
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Answer:

\omega_f=12.2954 \,rad.s^{-1}         i.e.

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Explanation:

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radius of the disk, R=0.2\,m

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radius of the chunk, r=0.04\,m

We know that the angular momentum is given by:

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4.9\times 0.2^2\times (2\times 2\pi)=(4.9\times 0.2^2+2.7\times 0.04^2 )\tiems \omega_f

\omega_f=12.2954 \,rad.s^{-1}

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3 years ago
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