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Naddika [18.5K]

2 years ago

6

The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block

is 110 m long.
Calculate the magnitude and direction of the displacement vector of path D. I cannot get the last part for direction

1 answer:

jarptica [38.1K]2 years ago

4 0

Answer: 592.37m

Explanation:

Person D is the blue line.

The total displacement is equal to the difference between the final position and the initial position, if the initial position is (0,0) we have that he first goes down two blocks, then right 6 blocks. then up 4 blocks, then left 1 block.

Now i will considerate that the positive x-axis is to the right and the positive y-axis is upwards.

Then the new position will be, if B is a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

And we know that B = 110m

P = (550m, 220m)

Now, then the displacement will be equal to the magnitude of our vector, (because the difference between P and the initial position is equal to P, as the initial position is (0,0)) this is:

P = √(550^2 + 220^2) = 592.37m

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1 year ago

Read 2 more answers

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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2 years ago

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alekssr [168]

Answer:

Explanation:

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2 years ago