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saw5 [17]
3 years ago
6

It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown

was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold?
Engineering
1 answer:
sertanlavr [38]3 years ago
4 0

Answer with Explanation:

The crown will be pure if it's specific gravity is 19.3

Now by definition of specific gravity it is the ratio between the weight of an object to the weight of water of equal volume

Since it is given that the weight of the crown is 11.8 N we need to find it's volume

Now According to Archimedes principle when the crown is immersed into water the water shall exert a force in upwards direction on the crown with a magnitude equaling to weight of the water displaced by the crown

Mathematically this is the difference between the weight of the crown in air and weight when immersed in water

Thus Buoyant force is F_{B}=11.8-10.9=0.9N

Now by Archimedes principle This force equals in magnitude to the weight of water of same volume as of the crown

Thus the specific gravity of the crown equals

S.G=\frac{11.8}{0.9}=13.11

As we see that the specific gravity of the crown material is less than that of pure gold hence we conclude that it is impure.  

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Answer: 10.29 sec.

Explanation:

Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.

If the car starts from rest, this means the following:

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100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec

Now, we calculate the change in energy:

ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J

<h2>If P= ΔK/Δt, </h2><h2>Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.</h2>
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3 years ago
Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician
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3 0
3 years ago
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

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AT INLET:

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h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

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\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

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Bingel [31]

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