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finlep [7]
3 years ago
9

A rigid copper tank, initially containing 1 m3 of air at 295k, 5 bar, is connected by a valve to a large supply line carrying ai

r at 295k, 15 bar. the valve is opened only as long as required to fill the tank with air to a pressure of 15 bar, at which point the air in the tank is at 310k. the copper tank, which has a mass of 20 kg, is at the same temperature as the air in the tank, both initially and finally. the specific heat of the copper is c = 0.385 kj/kg-k. assuming ideal gas behavior for the air, determine (a) the initial and final mass of air within the tank, each in kg, and (b) the heat transfer to the surroundings from the tank and its contents, in kj, ignoring kinetic and potential energy effects. (630)
Physics
1 answer:
professor190 [17]3 years ago
3 0
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A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th
DiKsa [7]

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

                        = 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= \frac{0.4937}{55\times10^{-6}}

Density of block = 8977.7 kg/m^3

4 0
3 years ago
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before
Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, d=2.5\times 10^{-15}\ m

Speed of light, c=3\times 10^{8}\ m\s

Let t is the time interval required for the strong interaction to occur. The speed is given by :

c=\dfrac{d}{t}

t=\dfrac{d}{c}

t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}

t=8.33\times 10^{-24}\ s

So, the time interval required for the strong interaction to occur is 8.33\times 10^{-24}\ s. Hence, this is the required solution.

8 0
3 years ago
In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0
3 years ago
A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T ·
Alex73 [517]

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, \phi=5\times 10^{-3}\ Tm^2

The relation between the self inductance and the magnetic flux is given by :

L=\dfrac{N\phi}{I}

L=\dfrac{17\times 5\times 10^{-3}}{4}

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

7 0
3 years ago
According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind
Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0
3 years ago
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