All categories

3 years ago

Connecting

Physics

1 answer:

svetoff [14.1K]3 years ago

4 0

5. what is the law called that states objects in motion must stay in motion/objects at rest will stay at rest?

You might be interested in

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

$h = 0.50 - 0.03 = 0.47m$

now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

$v = 3.03 m/s$

7 0

4 years ago

David rowed a boat upstream for three miles and then returned to point he started from. The entire journey took four hours. The

frez [133]

Upstream speed = S - 1
Downstream speed = S + 1

Average speed = total distance / total time

Average speed = (S - 1) + (S + 1) / 2
= S

S = 6 miles / 4 hours
S = 1.5 miles per hour

4 0

3 years ago

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

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$