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Mekhanik [1.2K]

3 years ago

7

I need help

1 answer:

valkas [14]3 years ago

6 0

Answer:

u make no sense can u just show a picture

Explanation:

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Rashid [163]

Answer:

1:knowledge of the structure of DNA

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7 0

3 years ago

The pilot of an airplane traveling 180km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The

gulaghasi [49]

I think it's 21 that's what I got when I was on that answer

4 0

3 years ago

there is a rusted nut we have a two spanners each of i 15 cm and 20 CM respectively which spanner is suitable to open the nut an

Novosadov [1.4K]

<h2>Answer : </h2>

20cm spanner is suitable to open this nut , Because we feel easy to apply force at the end of spanner to rusted nut. It means that as the perpendicular distance between force and axis of rotation increase the magnitude of moment also increase.

I think the Answer will helps you.....

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8 0

3 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

4 years ago

Your friend is flying a remote control jet in your yard, as seen in the diagram above. As it flies past you, you notice that the

rjkz [21]

Answer:c

Explanation:

7 0

4 years ago