Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is

Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e

- inversely proportional to its cross section area i.e

Therefore

ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:


......(1)
Again for tungsten:

........(2)
Given that
and 
Dividing the equation (1) and (2)

[since
and
]



Therefore the ratio of diameter of the copper to that of the tungsten is

The best and most correct answer among the choices provided by your question is the second choice.
<span>The major contributions of Maury included mapping the ocean bottom.</span>
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Answer:
1000 cm.
Explanation:
To obtain the estimated tree height :
(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)
Substituting values into the formula :
(150cm / 120 cm) = (height of tree / 800 cm)
Using cross multiplication :
Height of tree * 120 = 150 * 800
Height of tree = (150 * 800) / 120
Height of tree = 120,000 / 120
Height of tree = 1000
Hence, estimate height of tree = 1000 cm
Answer:

Explanation:
The inlet specific volume of air is given by:

The mass flow rates is expressed as:

The energy balance for the system can the be expresses in the rate form as:
![E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}](https://tex.z-dn.net/?f=E_%7Bin%7D-E_%7Bout%7D%3D%5Cbigtriangleup%20%5Cdot%20E%3D0%5C%5C%5C%5CE_%7Bin%7D%3DE_%7Bout%7D%5C%5C%5C%5C%5Cdot%20m%28h_1%2B0.5V_1%5E2%29%3D%5Cdot%20W_%7Bout%7D%2B%5Cdot%20m%28h_2%2B0.5V_2%5E2%29%2BQ_%7Bout%7D%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D%5Cdot%20m%28h_2-h_1%2B0.5%28V_2%5E2-V_1%5E2%29%29%3D-m%28%7Bcp%28T_2-t_1%29%2B0.5%28V_2%5E2-V_1%5E2%29%7D%29%5C%5C%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D-%2810.42lbm%2Fs%29%5B%280.25%5Cfrac%7BBtu%7D%7Blbm.%5Ctextdegree%20F%7D%29%28300-900%29%5Ctextdegree%20F%2B0.5%28%28700ft%2Fs%29%5E2-%28350ft%2Fs%29%5E2%29%28%5Cfrac%7B1%5Cfrac%7BBtu%7D%7Blbm%7D%7D%7B25037ft%5E2%2Fs%5E2%7D%29%5D%5C%5C%5C%5C%5C%5C%5C%5C%3D1486.5%5Cfrac%7BBtu%7D%7Bs%7D)
Hence, the mass flow rate of the air is 1486.5Btu/s