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Firlakuza [10]
3 years ago
15

Water can form large dewdrops in nature how would droplets made of vegetable oil instead of water be different

Physics
2 answers:
Jlenok [28]3 years ago
6 0

Answer:

d

Explanation:

>3

Natasha2012 [34]3 years ago
3 0

Answer:d

Explanation: oil would form droplets but only tiny ones because it’s surface tension is lower than that of water

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wolverine [178]

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alternating mountain ranges and valley.

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7 0
3 years ago
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Visible light waves do not diffract as well as radio waves because?​
Ivenika [448]
They have more energy than radio waves.
8 0
3 years ago
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a brand of a lightbulb has a power of 55w and an efficiency of 1.6% how much power is wasted as heat by this lightbulb
Maru [420]
98.4% of the power turns into heat.

If the bulb consumes 55 watts,
then it "wastes"

(0.984) x (55w) = 54.12 watts
5 0
3 years ago
An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The
Zolol [24]

Answer:

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

4 0
3 years ago
Ammonia enters the compressor of an industrial refrigeration plant at 2 bar, −10°C with a mass flow rate of 15 kg/min and is com
Juli2301 [7.4K]

Answer:

a) W=85.225 kW

b) 0.02\frac{kW}{K}

Explanation:

First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):

W+m*h_1=Q+m*h_2\\W=Q+m*(h_2-h_1)

Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, h_1=h(200kPa,-10C)=1440.6\frac{kJ}{kgK},  h_2=h(1200kPa,140C)=1757.5\frac{kJ}{kgK}

So, the power input to the compressor is:

W=6kW+15\frac{kg}{min}*\frac{min}{60s}*(1757.5-1440.6)\frac{kJ}{kg}\\W=85.225kW

b) The differential entropy change dS for a reversible heat transfer dQ at a temperature T is:

dS=\frac{dQ}{T}

This equation can be integrated if the heat transfer surface temperature remains constant, which is the case, giving as a result:

S_2-S_1=\frac{Q}{T}=\frac{6kW}{300K}=0.02\frac{kW}{K}

6 0
3 years ago
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