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Viefleur
1 day ago
8

Identify how scientists use radioactive isotopes by selecting the best answers from the drop-down menus. Archaeologists use radi

oactive isotopes to ___________ Scientists use radioactive isotopes in agriculture to_____________ Geologists use radioactive isotopes to ______________
Physics
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Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom
ValentinkaMS [3465]

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

D. Value of the centripetal force

3 0
3 months ago
A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
Ostrovityanka [3204]

Response:

\text{heat loss} = 24864.05 \ W/m^2

Clarification:

If

  • T_1, T_2 represent the temperatures of gases and liquids in Kelvins,
  • t_1 and t_2 denote the thicknesses of the gas layer and steel slab in meters,
  • h_1, h_2 are the convection coefficients for gas and liquid in W/m^2 \cdot K,
  • R_c represents the contact resistance in m^2 \cdot K/W,
  • and k_1, k_2 signify thermal conductivities of gas and steel in W/m \cdot K,

then: part(a):

\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

by employing known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Utilizing the rate equation:

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature is T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Correspondingly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature profile is depicted in the image provided

3 0
2 months ago
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