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zavuch27 [327]
3 years ago
7

A 30-kg child stands at one end of a floating 20-kg canoe that is 5.0-m long and initially at rest in the water. The child then

slowly walks to the other end of the canoe. How far does the canoe move in the water, assuming water friction is negligible
Physics
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

The canoe move 3 m in water.  

Explanation:

Given that,

Mass of the child, m = 30 kg

Mass of canoe, M = 20 kg

Length of canoe, L = 5 m

  • Since initially the boat is floating, its center of mass was at rest. So, when child moves on the boat from one end to the other end, the boat moves in opposite direction in such a way that the displacement of their center of mass remains at rest.            
  • Let x be the displacement of canoe in opposite direction to the motion of child, therefore  displacement of child is L - x

So, the displacement of the center of mass is given by :

d_{cm}=\dfrac{m(L-x)-Mx}{m+M}\\\\0=\dfrac{m(L-x)-Mx}{m+M}\\\\m(L-x)-Mx=0\\\\30(5-x)-20x=0\\\\x=3\ m

So, the canoe move 3 m in water.                                              

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Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
4 years ago
What do positive ions tend to do
notsponge [240]

Answer:

an electrically charged atom or group of atoms formed by the loss or gain of one or more electrons, as a cation (positive ion), which is created by electron loss and is attracted to the cathode in electrolysis, or as an anion (negative ion), which is created by an electron gain and is attracted to the anode.

7 0
3 years ago
If an engine at sea level produces 100 horsepower, how many horsepower would it develop at 6,000 feet of altitude?
Sonja [21]

Answer:

Power develop at 6000 ft = 100 - 18 = 82 hp

Given:

Power produced by the engine at sea level = 100 hp

Explanation:

For each 1000 ft rise above the sea level, the power loss of the engine is about 3%

Therefore,

1000 ft - 3% loss

6000 ft - 3\times 6 = 18 %

At 6000 ft, the power loss is 18% or 18 hp

Now,

Power develop at 6000 ft = Power at sea level - Power loss at 6000 ft

Power develop at 6000 ft = 100 - 18 = 82 hp

7 0
3 years ago
Which statement about these cells is correct
Nuetrik [128]
B is correct because most of the cells look the same so it's B
5 0
3 years ago
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after
s2008m [1.1K]

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

6 0
4 years ago
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