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2 years ago

9

1. What is the downward fores that acts on an airplane in flight?

1 answer:

Westkost [7]2 years ago

4 0

Answer: D. gravity

Explanation:

An airplane in flight is acted on by four forces: lift, the upward acting force; gravity, the downward acting force; thrust, the forward acting force; and drag, the backward acting force (also called wind resistance).

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Which of the following instruments is used to measure air pressure?

timama [110]

Barometer, it is used to measure air pressure

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3 years ago

The normal freezing point of a certain liquid

stepladder [879]

Answer : The molal freezing point depression constant of X is $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X= $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of X is $4.12^oC/m$

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3 years ago

The life function of transport in an organism

lesantik [10]

Answer:1

Explanation:

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3 years ago

If you pour water into a hot metal pan, the water will absorb heat energy from the pan. Will the water come to the same temperat

Marina CMI [18]

Answer:

Yes

Explanation:

It is heating up both

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2 years ago

Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C

taurus [48]

Answer:

$T_F=77.4\°C$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

$Q_1=-Q_2$

In terms of mass, specific heat and temperature change is:

$m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)$

Now, solve for the final temperature, as follows:

$T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}$

Then, plug in the masses, specific heat and temperatures to obtain:

$T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C$

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0

2 years ago