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igomit [66]
3 years ago
10

A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

ronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.
Physics
1 answer:
musickatia [10]3 years ago
7 0

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

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K=1400*V^2/2
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14*V^2=200*225
v^2=100*225/7
v=250/7^(1/2)

Answer: 250*7^(1/2)/7
4 0
3 years ago
A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.
PolarNik [594]

Answer:

11.95m/s

Explanation:

A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.

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Divide both sides by v²

m = 300/v² ..................  Equation 1

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Momentum = mv

25.1 = mv

Divide both sides by v

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Equate equations 1 and 2

300/v² = 25.1/v

Cross multiply

25.1v² = 300v

Multiply v with both sides

25.1v = 300

Divide both sides by 25.1

v = 300/25.1

V = 11.95m/s

I hope this was helpful, please mark as brainliest

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Answer:

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Given;

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Therefore, the final position of the ship after the given time period is  42 km West of B.

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Answer:

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Explanation:

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Answer:

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