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2 years ago

A 1200-kg car moving at 20 km/h is accelerated

Engineering

1 answer:

Grace [21]2 years ago

4 0

Answer:

Force on the car will be 4800 N and time required to cover this distance 13.75 sec

Explanation:

We have given mass of the car = 1200 kg

Initial velocity u = 20 km/h

Final velocity v = 75 km/h

Acceleration $a=4m/sec^2$

From the first equation of motion we know that

v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So $75=20+4\times t$

t = 13.75 sec

From second law of motion we know that $F=ma$

So force $F=1200\times 4=4800N$

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1 year ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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3 years ago

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lyudmila [28]

Answer:

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Explanation:

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2 years ago

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3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$