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3 years ago

What happens when the temperature of an object decreases

Physics

1 answer:

Helen [10]3 years ago

5 0

The molecules become more stiff and move slowly

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Volcanic eruptions are natural phenomena in the . They sometimes erupt, spilling out lava and dust. The lava solidifies, forming

solniwko [45]

Answer:

B. Geosphere

A. Biosphere

A. Atmosphere

Explanation:

Volcanic eruptions occurs within the Geosphere. The geosphere is the rock solid earth make up of rocks that extends into the deep interior.

Magma formed deep within the crust rises to elevated parts and finally erupts as lava on the surface. When they cool, they solidify to form volcanic rocks.

The volcanic eruptions affects the biosphere significantly. The biosphere is the portion of the earth where all life forms exists.

Gases and ash spewed during an eruption into the atmosphere causes severe changes to weather and leads to pollution. The atmosphere is the gaseous envelope round the earth.

3 0

3 years ago

A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m,

timurjin [86]

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance <em>d</em> underwater would be:

$W=F.d=-Fd$

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

$W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)$

Putting all together:

$F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N$

7 0

3 years ago

A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight

LuckyWell [14K]

Answer:

At an angle of $30^{\circ}$

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

$R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}$

6 0

2 years ago

Which ball (if either) has the greatest speed at the moment of impact

Ierofanga [76]

Are there any options??

I would have to say metal of course but without options I can't assume anything

5 0

3 years ago

Read 2 more answers

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

5 0

3 years ago