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Ann [662]
3 years ago
14

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

ity of 5.21 m/s relative to the shore. Find the recoil velocity that the raft would have (again relative to the shore) if there were no friction and resistance due to the water.
Physics
1 answer:
nalin [4]3 years ago
7 0

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

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Answer:

28.79%

Explanation:

Given

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Using the following formula, we'll solve the required rate of superelevation.

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4 years ago
You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,
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Answer:

The value is T_c  =  12 .1 ^oC

Explanation:

From the question we are told that

The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

The temperature of the ice cube is T_i = 0^o C

The mass of the copper cube is m_c  =  0.540 \  kg

The final temperature of both substance is T_f  =  0^oC

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

Q =  m_c  *  c_c *  [T_c  -  T_f ]

The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

Generally the heat gained by the ice cube is mathematically represented as

Q_1 =  m_i * L

Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

So

Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

=> Q_1 =  2505 \ J

So

2505  =  0.540  *  385 *  [T_c  - 0 ]

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