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yarga [219]
2 years ago
8

A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and

110KN (Compression) on the 70mm by 10mm face Determine the force to be applied on the 70mm by 50mm face such that there is no change in volume
Physics
1 answer:
sammy [17]2 years ago
7 0

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

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alukav5142 [94]

Answer: 0.258

Explanation:

The resistance R of a wire is calculated by the following formula:

R=\rho\frac{l}{s}    (1)

Where:

\rho is the resistivity of the material the wire is made of. For aluminium is \rho_{Al}=2.65(10)^{-8}m\Omega  and for copper is \rho_{Cu}=1.68(10)^{-8}m\Omega

l is the length of the wire, which in the case of aluminium is l_{Al}=12m, and in the case of copper is l_{Cu}=30m

s is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

s=\pi{(\frac{d}{2})}^{2}  (2) Where d  is the diameter of the circumference.

For aluminium wire the diameter is  d_{Al}=2.5mm=0.0025m  and for copper is d_{Cu}=1.6mm=0.0016m

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}

s_{Al}=0.000004908m^{2}   (3)

<u>For copper:</u>

s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}

s_{Cu}=0.00000201m^{2}    (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}     (5)

R_{Al}=0.0647\Omega     (6)  Resistance of aluminium wire

<u>Copper wire:</u>

R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}     (6)

R_{Cu}=0.250\Omega     (7)  Resistance of copper wire

At this point we are able to calculate the  ratio of the resistance of both wires:

Ratio=\frac{R_{Al}}{R_{Cu}}   (8)

\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}   (9)

Finally:

\frac{R_{Al}}{R_{Cu}}=0.258  This is the ratio

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Answer:

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Explanation:

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<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

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