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Hatshy [7]
2 years ago
14

A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled

out of the window. Which law, principle, or equation can be used to explain this movement of the curtain?A. Poiseuille's lawB. Bernoulli's equationC. the equation of continuityD. Archimedes' principleE. Pascal's principle
Physics
1 answer:
VikaD [51]2 years ago
3 0

Answer:

option B.

Explanation:

The correct answer is option B.

The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

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Which of the following statements best represents the impact of evolutionary theory on the field of psychology?
grandymaker [24]

Answer- the correct option for the question would be choice number c which tells us about the natural selection process of human species.

Explanation- Evolution of psychology was studied under the given heads namely Behaviorism cognitive functionalism sociocultural Psychodynamic and structuralism.  

On studying the facts under these prospects the evolution of psychology gives us the idea of the selection by nature of individuals and the pattern of their behavior shown in different aspects.

8 0
2 years ago
An electric toaster is rated 1200 watts at 120 volts. what is the total electrical energy used to operate the toaster for 30 sec
Anarel [89]
Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.

If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.
3 0
2 years ago
An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

4 0
2 years ago
From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco
Vanyuwa [196]

Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

Explanation:

Given that,

From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.

At the highest point,

mgh+\dfrac{1}{2}mu^2=mgh'+0          

100=mgh'

At lowest point,

mgh'=K

From above two equation, we get :

Kinetic energy, K = 100 J

So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).                                                                        

7 0
2 years ago
If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        a = \frac{u *  \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}

Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

=>   \frac{\Delta m}{\Delta t}   = \frac{170100}{2027}

=>   \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

     

3 0
2 years ago
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