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3 years ago

Which type of heat transfer takes place in gases by the movement of particles through a medium?

Physics

2 answers:

Vitek1552 [10]3 years ago

6 0

( Convection ) is the movement of the heat by a fluid such as water or air.

Sergeu [11.5K]3 years ago

4 0

Answer: Convention

Explanation: Convention is pretty well known for being a process that transmits heat from one place to another place with the movements of heated particles. I got this answer from my notebook during my chemistry class.

Hope this answer helps!

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In Florida, most of the rocks that are underground are a type of soft, white rock called limestone. If limestone is underwater f

nadya68 [22]

D. Physical weathering

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3 years ago

Read 2 more answers

Two cars of the same mass collide at an intersection. Just before the collision, one car is traveling east at 80.0 km/h and the

Elden [556K]

Answer:

$v_{f}=70\frac{km}{h}$

Explanation:

In a completely inelastic collision the momentum P is conserved, so can be expressed as:

$P_{f}-P_{i}=0$

As P is defined as: $P=m.v$

Replacing:

$m_{1}.v_{1f}+m_{2}.v_{2f}=m_{1}.v_{1i}+m_{2}.v_{2i}$ (Eq. 1)

As the problem says that the two cars have the same mass:

$m_{1}=m_{2}=m$

And in a completely inelastic collision the velocities after the collision are equal, so:

$v_{1f}=v_{2f}=v_{f}$

So replacing in Eq. 1:

$m.v_{f}+m.v_{f}=m.v_{1i}+m.v_{2i}$

$2m.v_{f}=m.(v_{1i}+v_{2i})$

Solving for $v_{f}$ :

$v_{f}=\frac{(v_{1i}+v_{2i})}{2}$

And replacing the values for the velocity:

$v_{f}=\frac{(80.0\frac{km}{h}+60.0\frac{km}{h})}{2}$

$v_{f}=70\frac{km}{h}$

5 0

3 years ago

A member of the Penn State Women's Basketball team performs a jumping movement simulating that used in a jump shot. The time fro

Shkiper50 [21]

A) Starting from rest, we have the entire cycle determined by 0.8s.

If we assume a constant movement, half of that time is when it reaches the highest point, that is, in 0.4s.

The distance as a function of speed and acceleration is given by,

$d=v_it+\frac{1}{2}at^2$

At the initial point the speed is zero and the acceleration is equivalent to gravity.

$d= 0+ \frac{1}{2}9.8*0.4^2$

$d=0.784m$

B) When returning to the ground, the final speed is zero. Therefore, the equation that relates velocity to acceleration is given by,

$V_f = V_i+at$

$0 = V_i -9.8(0.4)$

$V_i = 3.92m/s$

5 0

4 years ago

Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to

Nataliya [291]

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

$P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1} =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}$

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

$P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}$

As we know, P1 must be equal to $\frac{F_{1} }{A_{1}}$ , so, replacing P1 in the equation, we have:

$P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})$

And

$F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))$

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as $Q = V A$ , where Q is flow, V is velocity and A is area. As the same, we can write this: $Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}$ . In the last two equations, let's clear Velocities.

$V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}$

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

$F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))$

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

$D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg$

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

3 0

3 years ago

Which set of equipment would be most useful to determine the density of a liquid?

NemiM [27]

Triple beam balance/scale to find mass, and graduated cylinder to find volume

3 0

4 years ago