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bezimeni [28]
3 years ago
12

Convection is a mode of heat transfer for which types of objects? A. solids and liquids B. gases and solids C. gases only D. liq

uids and gases E. liquids only.
Physics
2 answers:
Airida [17]3 years ago
7 0

The correct answer to the question will be D). Liquids and gases.

EXPLANATION:

Convection is the type of mode of heat transfer in which there will be actual motion of particles from one part to the another part of a fluid.

In case of liquids, the particles which are present at the bottom of the container get heated up first and become lighter. The lighter particles will go up and the top heavier particles will move downward. Again the same process will be repeated. In this way, a convection cycle is produced.

For instance, heating of water.

In case of gases, same thing happens just like liquid. The air particles will be heated up first and becomes lighter. These heated particles will move upward. Due to this, an empty space will be created in that region. In order to occupy the empty space, the air will flow from another region which is at low temperature as compared to the heated region. Again, the same process will be repeated which will results into the formation of convection cycle .

Hence, convection occurs both in liquids and gases.



OLga [1]3 years ago
7 0
The heat transfer involving convection is a process that is dependent in the density gradients of substances. This is better suited for fluids which are composed of the liquids and gases. Hence, the answer for this item is most appropriately letter D. 
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the idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse
Finger [1]

Forward thrust has positive values and reverse thrust has negative values.

Thrust is a sudden push or pull in a certain direction.

a)

Flight speed u = 150 km/h

1 km/h = \frac{1}{3.6} km/s

therefore, 150 km/h =  41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

T = m_{exhaust} * U_{exhaust} - U_{flight}

T = 50 * (150 - 41.67)

T = 5416.67 N

Therefore, the value of forward thrust is 5416.67 N.

b)

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,

thus thrust equation is:

T = m_{exhaust} * U_{exhaust} - U_{flight}

T = 50 * (0 - 41.67)

T = -2083.5 N

Therefore, the thrust force T is -2083.5 N in the reverse direction.

c)

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero becauseU_{exhaust} = U_{flight} = 0\\

T = 0

Therefore, there is no difference in two velocities in x direction.

The given question is incomplete, the complete question is,

"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. What is the forward thrust of this engine?

b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?

c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"

To know more about thrust,

brainly.com/question/14552836

#SPJ1

3 0
1 year ago
A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure
klasskru [66]

Answer:

3x_0

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

d=v_x t

where

v_x is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

d=v_0 t = x_0

in the second case, the horizontal velocity is increased to

v_x' = 3v_0

And so the new distance travelled will be

d' = v_x' t = 3 v_0 t = 3 x_0

So, the distance increases linearly with the horizontal velocity.

5 0
3 years ago
Doubling the frequency of a wave source doubles the speed of the wave assuing that wavelength remains the same? True or false? W
SVETLANKA909090 [29]

Answer:true,because velocity is directly proportional to speed or velocity

Explanation:

Velocity = frequency x wavelength

The velocity or speed varies directly with the frequency, so as the frequency is increased, the velocity or speed is also increased

8 0
3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

also

F\propto \frac{1}{dt}

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

I=F.dt

I=m.a.dt

I=m.\frac{dv}{dt}.dt

I=m.dv=dp

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0
2 years ago
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