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3 years ago

What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Physics

1 answer:

12345 [234]3 years ago

5 0

Incomplete question.The complete question is

A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Answer:

Total gauge pressure= 3724Pa or 3.724kPa

Explanation:

Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.

Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.

Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa

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A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b

Andru [333]

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390 J

ΔU= 330 J

5 0

2 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

2 years ago

Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65

guajiro [1.7K]

Answer:

She must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, $\theta =65$ °

Range of the projectile is, $R=520$ m.

Acceleration due to gravity, $g=9.8\ m/s^2$

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

$R=\frac{v_{0}^2\sin2\theta}{g}$

Plug in all the given values and solve for minimum speed, $v_0$ .

$520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}$

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0

3 years ago

How does reducing the mass of a moving object by half (1/2) change its kinetic energy?

allochka39001 [22]

Kinetic energy = 1/2 m v²
If we reduce the mass by half > m/2
Kinetic energy = 1/2 m/2 v²
We should know that 1/2 × 1/2 = 1/4
So kinetic energy will be :
1/4 × m × v²

7 0

3 years ago

An AA battery is connected to a parallel-plate capacitor having 3.9cm×3.9cm plates spaced 1.5 mm apart. How much charge does the

djyliett [7]

Answer:

Q = 1.35*10⁻¹¹ C.

Explanation:

By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:

$C= \frac{Q}{V}$

At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:

C = ε₀*A / d

Replacing by the values of A, and d, and taking into account that

ε₀ = 8.85*10⁻¹² F/m,

we get the value of the capacitance as follows:

C = 8.97*10⁻¹² F

As the voltage of an AA battery is 1.5 V, and is all applied to the capacitor, we can conclude that the charge on one of the plates is as follows:

Q = C* V = 8.97*10⁻¹² F* 1.5 V = 1.35*10⁻¹¹ C

5 0

2 years ago

Read 2 more answers