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2 years ago

Why does liquid water boil away when exposed to the Martian atmosphere?

Physics

1 answer:

Studentka2010 [4]2 years ago

6 0

Answer: d. the low air pressure lowers the boiling point

Explanation:

Note that there cannot be less than 6.1 millibars of liquid water. This fact is important because the Martian surface atmospheric pressure hovers just below that value. Any water that could form from melting water on a warm afternoon will easily vanish into the desiccated Martian atmosphere.

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Large amplitude of sound vibrations will produce.....

user100 [1]

Answer:

louder sound.

Explanation:

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound

8 0

3 years ago

What is the volume of a storage tank which will hold 3200kg of petrol?

otez555 [7]

Answer:

The volume of 3200 kg of petrol is 4 m^3.

4 0

2 years ago

When an oxygen atom forms an ion, it gains two electrons

kirill [66]

Note the atom of the Oxygen is electrically neutral, meaning it has equal numbers of electrons and protons.

So if it gains 2 electrons, it would have excess of 2 electrons, hence its charge would be -2.

Option B.

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2 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

4 0

2 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$