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devlian [24]
3 years ago
11

A coffee pot in the form of a circular cylinder of radius 3 in. is being filled with water flowing at a constant rate. if the wa

ter level is rising at the rate of 0.4 in./sec, what is the rate at which water is flowing into the coffee pot? (round your answer to two decimal places.)
Physics
1 answer:
Ad libitum [116K]3 years ago
6 0
This is a question volumetric flow rate.
In 1 second, water rises with 0.4 in.

Additionally,
Volume filled in 1 second = A*0.4, A = surface area of the circular cylinder.

Where,
A = πR^2 = π*3^2 = 9π in^2

Therefore,
Volumetric flow rate = 9π*0.4 = 11.31 in^3/s

In terms of liters/s,
1 in^3 = 0.0163871 litres
Then,
11.31 in^3/s = 11.31*0.0163871 ≈ 0.19 l/s
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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s

Explanation:

a. let m_A be the mass of block A, andm_B=10.0kg be the mass of block B. The initial velocity of A,\rightarrow v_A_1=2.0m/s

-The initial momentum =Final momentum since there's no external net forces.

pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

v_A_1-v_B_1=v_{B2}-v_{A2}

-Applying the conservation of momentum. The blocks have the same velocity after collision:

v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s

#Total Mechanical energy before and after the elastic collision is equal:

K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s

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