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mafiozo [28]
3 years ago
9

Two friends are standing on opposite ends of a canoe. The canoe is at rest with respect to the lake. Then the friend at the nort

h end of the canoe throws a very massive ball to the friend at the south end of the canoe, who catches it. How does the canoe move, if at all, as a result? A. The canoe accelerates to the south. B.The canoe remains motionless. C. The canoe accelerates to the north
Physics
1 answer:
juin [17]3 years ago
7 0
-- When the man at the North end throws the ball, the canoe
accelerates to the North.

-- While the ball is in flight, traveling south by the length of the
canoe, the canoe moves northward at a constant speed.

-- If the man at the south end misses the ball and fails to catch it,
then the canoe continues moving south at a constant speed, even
after the ball has passed him.

-- If the man at the south end successfully catches the ball, then the
canoe accelerates south at that moment, and comes to rest in the water.
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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
2 years ago
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the
san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

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3 years ago
What type of friction is using chalk in the summer to draw on the ground in Copley square?
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The second option rolling friction
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An atom of sodium (Na) gives up an electron to chlorine (Cl) to form table salt. What type of bond is this?
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Your answer will be C!~
6 0
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Doubling an object’s height will have what effect on its potential energy due to gravity?
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Potential energy due to gravity = Ep = mgh [symbols have their usual meaning ]
Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.
5 0
2 years ago
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