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Reika [66]
3 years ago
10

An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 1.3 cm/s at an angle of 61◦ with the lon

g edge. How long will it take her to get across? Answer in units of s.
Engineering
1 answer:
ankoles [38]3 years ago
7 0

Answer:

t = 30.1 sec

Explanation:

If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.

If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.

We are only interested in the component of the velocity across the paper, that can be calculated as follows:

vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.

We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:

vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s

Applying the definition of average velocity, we can solve for t:

t =\frac{x}{vx} = \frac{34.2 cm}{1.14 cm/s} =30.1 sec

⇒ t = 30.1 sec

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The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%
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and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

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Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0
3 years ago
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