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Degger [83]
3 years ago
10

Cylindrical rod has equal and opposite forces applied perpendicular to its circular ends. The forces are directed away from the

rod, stretching the rod. What type of stress is this rod subject to:_________A) The rod is subject to shear stress. B) The rod is subject to tensile stress. C) The rod is subject to compressive stress. D) The rod is subject to bulk stress.
Physics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

B) The rod is subject to tensile stress

Explanation:

Tensile stress is the force induced in the body per unit area in response to externally applied force, which stretches the body.

Tensile stress is attached with tensile forces. It is makes material elongate along the axis of the applied load.

It is given as magnitude (F) of the force that is applied along an elastic rod divided by the cross sectional area (A) of the rod in a direction that is perpendicular to the applied force.

materials such as Doctile have the ability to withstand the load while brittle materials who not have the ability they fail before reaching the ultimate material strength.

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Answer:

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2 years ago
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A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

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2 years ago
G a person of mass 100 kg is riding an elevator which was initially moving up with a velocity of 3 m/s. over a distance of 4 m t
andriy [413]
E=mc² where c is speed of the light
3 m/s more andmore less than speed of the light. So mass of the person still 100 kg
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What has been something positive for you that has been a result of social distancing? (Answer thoughtfully and throughly.)
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Answer: I have more time to think about the wonderful things in life that we don't always appreciate.

Explanation:

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For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

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