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3 years ago

12. A cross-country race car driver sets out on a 1.00 hour, 100.0 km race. At the halfway marker

Physics

1 answer:

Akimi4 [234]3 years ago

8 0

Answer:

A.) Time = 0.625 hrs = 37.5 minutes

B.) Speed required = 133.33km/hr

Explanation:

Given the following :

Total race distance = 100km

Total time required = 1 hour = 60 minutes

Average speed after traveling 50km = 80km/hr

80km/hr : This means it will take one hour to cover a distance of 80km

Therefore, time taken to cove first 50km at that average speed equals :

1 hour = 80 km

t hours = 50km

80t = 50

t = 50/80

t = 5/8 hours

t = 0.625hours

t = 0.625 * 60 = 37.5minutes

Average speed required to complete the race in 1 hour = 100km/hr

Time used to complete first 50km = 0.625 hour

Time remaining: (1 - 0.625) hour = 0.375

Speed required = Distance left / time left

Speed required = 50 / 0.375

Speed required = 133.33km/hr

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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 $\sigma_s = - 1795.84 psi$

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