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Bess [88]
3 years ago
14

Which of the following devices can measure only atmospheric pressure? A. Pressure gauge B. Mercury tube C. Tire gauge D. Baromet

er
Chemistry
1 answer:
kap26 [50]3 years ago
4 0
D. Barometer. Hope this helps
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Solve the problem. Express your answer to the correct number of significant figures
Shalnov [3]

Answer:

Explanation:

30.53131

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3 years ago
3. Which term is used to describe the highest point on a transverse wave?
ddd [48]
The highest point is the crest
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3 years ago
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How the inside of a greenhouse works is an example of conduction, convection, or radiation?
gizmo_the_mogwai [7]

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Radiation

Explanation: Because greenhouses have glass and sun travels through glass .

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3 years ago
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Thermohaline currents are influenced by wind.<br><br> True or false?
tresset_1 [31]

Answer:

FALSE

Explanation:

The under-water deep ocean currents that are generated due to the differences in the water density are generally termed as the thermohaline circulation. It is mainly controlled by the two factors namely the temperature and the salinity. The word 'thermohaline' is directly derived from the temperature (thermo) and salinity (haline).

This thermohaline circulation is considered as the oceanic conveyor belt that allows the water to move under the surface of water at certain depths from the equator to the poles and back from the poles to the equator.

Thus, it is directly associated with the density of the water. It has no relation with the wind.

Hence, the above statement is False.

6 0
3 years ago
A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128
QveST [7]

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

DH_{lead}=-DH_{water}

Now, by stating the heat capacity definition:

m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\

Solving for the equilibrium temperature:

T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.

5 0
2 years ago
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