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Rashid [163]
4 years ago
12

A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d

egrees south of that direction, and to continue to drive 50 miles deviating by 30 degrees each time until he returns to his original position. How far will he drive, and how many vectors must he sum to calculate his displacement
Physics
1 answer:
stira [4]4 years ago
5 0

Answer:600 miles, 12

Explanation: The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.

The sum of exterior angles of a polygon = 360 degrees.

Exterior angle of a polygon = (360 ÷ number of sides)

Therefore,

Number of sides = 360 ÷ exterior angle

Exterior angle = 30 degrees

Hence,

Number of sides = 360 ÷ 30 = 12 sides

Since distance traveled of 50 miles is the same for each displacement ;

Total displacement = distance traveled * number of sides

Total displacement = 50 * 12 = 600 miles.

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

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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\\\\

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 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}

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3 years ago
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3 years ago
A gas, behaving ideally, fills a fixed volume container at a pressure P₁ and at a temperature T₁. The temperature of the contain
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Answer:

c. P₁/T₁=P₂/T₂

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the volume volume is said to be filled, i.e its is kept constants when temperature is change

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