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Ne4ueva [31]
3 years ago
13

What’s a educated guess supported by evidence?

Physics
2 answers:
Sauron [17]3 years ago
4 0

a hypothesis is an educated guess :)

worty [1.4K]3 years ago
3 0
An educated guess is like an estimate except instead of completing a problem you use other variables (such as familiar numbers or simplified answers) to answer the question. An educated guess can be supported by evidence by using the parts of a you used to come up with the guess in your answer. I believe 2^1 is equal to 2 because any number raiser to the power of 1 is equivalent to one. The because is you evidence and the answer is your educated guess. Hope this helped!
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An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
3 years ago
Given a sphere with radius r, find the height of a pyramid of minimum volume whose base is a square and whose base and triangula
BaLLatris [955]

Answer:

The Height of a pyramid is 4r.

Explanation:

Explanation is in the following attachments

6 0
3 years ago
Two resistors, R1 and R2, are
dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)

1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)

<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0
3 years ago
How is energy transferred during the water cycle? Question 1 options: Water gains energy during evaporation and releases it duri
Margaret [11]

Answer:

Water gains energy during evaporation and releases it during condensation in the atmosphere

Explanation:

In the water cycle, heat energy is gained or lost by water as it undergoes various processes in the cycle.

In evaporation, water molecules gains energy because the molecules of water vibrate faster and become more energetic. Hence they are able to escape into the atmosphere from the surface of the liquid.

In condensation, the molecules of gaseous water looses energy and becomes liquid.

Hence, water gains energy during evaporation and releases it during condensation in the atmosphere.

8 0
2 years ago
Read 2 more answers
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