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serg [7]
3 years ago
13

Wind resistance _____ as speed increases.

Physics
2 answers:
Rudik [331]3 years ago
7 0

The correct option is option B that wind resistance increases dramatically.

Explanation:

When the speed of an object increases resistance increases drastically because the object pushing through air has obtained a high speed but the air is unable to move out of the way so fast so show it tends to get compressed and oppose the moving object and the object names to put more pressure because some of the pressure goes into cutting air and moving through it.

Irina18 [472]3 years ago
6 0

Answer:

B. increases dramatically

Explanation:

Wind resistance increases dramatically as the speed increases.

Resistance tends to prevent the motion of a body. When speed increases, the particles of air are set into motion and this energizes them. More air gains increased acceleration and they collide with a body. This initiates a resistance against the motion of such body. Therefore, wind resistance increases rapidly as speed increases.

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¿Cuál de las siguientes magnitudes es derivada?
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Answer:

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Explanation:

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3 0
3 years ago
Crystals are made using supersaturated solutions of solids in water.
labwork [276]

The way these supersaturated solutions are made is: A. The water would need to be heated to a higher temperature, which would give molecules and ions more  kinetic energy, increasing solubility.

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Generally, supersaturated solutions of solids in water are typically used for the creation of crystals because they are able to hold more of the solute than they would at room  temperature.

In order to create these supersaturated solutions, the water should be heated to a higher temperature, so that the water molecules and ions can gain more kinetic energy and thereby increasing solubility.

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Read more: brainly.com/question/24058779

8 0
3 years ago
Read 2 more answers
A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i
AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume v and enthalpy h as,

h_1 = 95.96Btu/lb (  h_1 = h_f at 2psia )

v_1 = 0.016238ft^3/lb ( v_1 = v_f at 2psia )

w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb

w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb

h_3 = 1364.0Btu/lb

s_3 = 1.5073Btu/lb.R

( at P_3 = 1500psia & T_3 = 800^0F )

P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765

( S_f & S_{fg} when pressure is 2psia)

h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb

n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb

Therefore,

q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb

To calculate the mass flow rate of steam in the cycle, we use the formula

W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s

where 1Kj = 0.947817 Btu

The power output and the rate of heat addition are calculated thus,

W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW

Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s

The thermal efficiency of the cycle can be found thus;

n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343

= 34.3%

5 0
3 years ago
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