What advice would you give her to improve her diagram? Choose all that apply.

Physics

2 answers:

Ymorist [56]2 years ago

5 0

Answer:

A, C

Explanation:

In order for a box to be moving at constant velocity, that means acceleration must be 0(as the speed isn't increasing). Thus, since F = ma, and a is 0, the net force must also be 0. The diagram must be completely balanced in every direction, which is why the two horizontal arrows must be equal values.

C is also true because gravity is a stronger force here than the applied force. We can calculate the force due to gravity with Fgravity = mg. Thus, we get the force as a result of gravity = 9.8*50, which is around 500N and much larger than the applied force of 200N.

Lisa [10]2 years ago

3 0

I think B but I would need to see how much forces are applied

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Calculate the root mean square velocity of nitrogen molecules at 25 C.

iragen [17]

root mean square<span>= square root of ( 3RT/M)

R = 8.314 J/K/mole
T = 25 + 273 = 298 K
M = molecular mas of N2 in kg = 28 X 10^-3 kg

put values...

</span><span> root mean square</span> = square root of ( 3 X 8.314 X 298/28 X 10^-3)
= square root of ( 265454.143)
= 515.2 m/s
so option A is right
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4 0

3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data: