Need some help with my home work.

In the situation shown below, what would the Moon look like from Earth? Sun, Earth and Moon Four Moon Views A. View A B. View B

Blababa [14]

Answer:

where is the picture?

Explanation:

3 0

3 years ago

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

5 0

3 years ago

If you log into the admin account on windows 10, will the admin be notified ?

professor190 [17]

Answer:

Just message the Admin,

Ok.

6 0

3 years ago

Read 2 more answers

An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,

blsea [12.9K]

Answer:

the % recovery of aluminum product is 80.5%

the % purity of the aluminum product is 54.7%

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = 80.5%

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = 54.7%

8 0

3 years ago

The time delay of a long-distance call can be determined by multiplying a small fixed constant by the number of communication li

aliina [53]

Answer:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

Remove all leaves of T. Let the remaining tree be T1.
Remove all leaves of T1. Let the remaining tree be T2.
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
If Tk has only one node, that is the center of T. The diameter of T is 2k.
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.

Explanation:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

Remove all leaves of T. Let the remaining tree be T1.
Remove all leaves of T1. Let the remaining tree be T2.
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
If Tk has only one node, that is the center of T. The diameter of T is 2k.
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.

4 0

3 years ago