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3 years ago

Is it possible for speed to be constant while acceleration is not zero?

1 answer:

5 0

Sure enough !

Acceleration means any change in speed OR DIRECTION of motion.

So if something is moving in a circle or around a curve, even at constant
speed, its direction is changing, so acceleration is not zero.

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A 2.4-kg cart is rolling along a frictionless, horizontal track towards a 1.7-kg cart that is held initially at rest. The carts

inessss [21]

Answer:

Explanation:

Mass of first cart M1=2.4kg

Velocity of first cart U1=4.1m/s

Mass of second cart M2=1.7kg

Second cart is initially at rest U2=0

After an instant, the velocity of the second cart is U2=-2.8m/s

Now after collision the two cart move together with the same velocity I.e inelastic collision

Using conservation of momentum

Momentum before collision, = momentum after collision

M1U1 + M2U2 = (M1+M2)V

2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V

9.84 - 4.76 = 4.1V

5.08=4.1V

V=5.08/4.1

V=1.24m/s

The momentum of the two cart at that instant is

M1U1+M2U2

2.4×4.1 + 1.7× -2.8

9.84 - 4.76

5.08kgm/s

So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s

3 0

3 years ago

_______________have a negative charge and are located on the outside of the nucleus.

GuDViN [60]

May 5th is Adachi day change your profile picture

6 0

4 years ago

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:

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