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Butoxors [25]
3 years ago
10

As a pendulum moves closer to the equilibrium position, how do the velocity, acceleration, and force change? The velocity decrea

ses, the acceleration decreases, and the net force decreases. The velocity increases, the acceleration decreases, and the net force decreases. The velocity increases, the acceleration increases, and the net force decreases. The velocity decreases, the acceleration increases, and the net force increases.
Physics
2 answers:
alexandr1967 [171]3 years ago
4 0

Answer:

The velocity increases, the acceleration decreases, and the net force decreases.

Explanation:

Velocity is at maximum when acceleration is zero thus as the pendulum moves closer to equilibrium position, velocity increases and acceleration decreases. As the pendulum moves closer to equilibrium, the magnitude of force decreases until it reaches the equilibrium position when the magnitude of force is zero. This means the net force decreases in the same manner the acceleration decreases because from the formula of force which is the product of mass and acceleration.

anzhelika [568]3 years ago
3 0

Answer:

B

Explanation:

on edge

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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
A ball was thrown from a projectile building of 30m which moves at a constant velocity of 20m/s and has an angle of 30degrees to
qwelly [4]

Answer:

Time of flight = 4.08seconds

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Explanation: complete question( and the horizontal component of the initial velocity.)

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T=( 2×20)/9.8

T= 40/9.8= 4.08seconds

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6 0
3 years ago
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5 0
4 years ago
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As the headwind comes at the head of the plane, therefore 50 mph headwind slows the plane down.

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time = \frac{1250 mil}{500 mph}

time = 2.5 hour

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5 0
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vredina [299]
The correct answer is c. Medium!
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