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3 years ago

What is the difference in the meaning between the lengths of lines in a phylogram and ultrametric tree?

1 answer:

7 0

<span>A phylogram is a phylogenetic tree where edge lengths signify time or genetic distance while an ultrametric tree or “chronogram” is a phylogenetic tree where edge lengths represent time. Also i<span>n an ultrametric tree, all tips of the tree are equidistant from the root of the tree. While a phylogram has dissimilar branch lengths which quantitatively signify the opposing degrees of change between the root and each terminal node.</span></span>

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Two identical pebbles are dropped. The first is dropped from a height of 256 feet and the second is dropped from a height of 400

ehidna [41]

Answer:

4.022 seconds and 4.99 seconds

Explanation:

Hello!

The free fall of the stone corresponds to a uniformly varied rectilinear movement

d=V_0*t+1/2*g*t^2

Being a free fall the initial speed is zero.

The distance is positive when considered in the same direction and direction as acceleration and speed.

256 feet stone

79.25 m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t = 4.022 seconds

400 feet stone

121.92m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t= 4,99 seconds

success with your homework!

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8 0

3 years ago

"The toy car is about 3 inches long" is an example of a what observation?

Crazy boy [7]

Answer:

It's a quantitative observation because it includes numerical data.

Explanation:

please mark brainliest if im correct

5 0

4 years ago

Describe briefly the potential energy it has ( roller coaster)

jek_recluse [69]

Answer:

The energy that it starts with at the top of the inclinede Plane

Explanation:

Hope I helped:)

5 0

3 years ago

What is the force of friction acting on the crate? I am offering 30 points.

olganol [36]

I think the answer is C

4 0

3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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3 years ago