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2 years ago

6

Sorry I keep asking questions, but if two forces on an object are balanced, will it move? (Im also trying to get rid of my point

s) XD

2 answers:

Dmitrij [34]2 years ago

6 0

If they are equal amounts in force then No it won't.

muminat2 years ago

4 0

If the two forces are acting on the object in opposite directions(ex: push from the back and a push from the front), the object will not move.

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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

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

2 years ago

Which of the following is a way in which earth material is returned to the surface?

kvv77 [185]

Answer:

Explanation:

soil

4 0

3 years ago

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if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

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2 years ago

150J of heat energy

arsen [322]

Btu/(lb-°F) J/(g-°C i mean this is the correct answer

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1 year ago

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A radio-controlled car increases its kinetic energy from 4 j to 12 j over a distance of 2 m. what was the average net force on t

just olya [345]

The answer to this is easy once you look at the units for Joules. 1 Joule = 1 N.m (Newton.meter). The 'Newton' is the units of force that we are trying to find, and we know the meters is 2, from the question. So you have an 8Joule or 8N.m energy difference over 2 meters.

well if we know the meters, then the real question is written as:
8N.m = ?N x 2m
so just solve for N;
N = 8N.m / 2m = 4
So F = 4N

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3 years ago