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3 years ago

The coordinates 40 degrees north and 100 degrees east falls in what country?

2 answers:

7 0

We can find for the country by using a map with angle coordinates. Based on the map that I found online, the country which is located at around 40 degrees north and 100 degrees east is no other than:

<span>China</span>

Ipatiy [6.2K]3 years ago

5 0

Answer:

China

Explanation:

Using the map one can find out this. The location lies in the northern China. It lies in the northern Chinese province Inner Mongolia.

The coordinate 40 degrees north refers to the latitude of the place and 100 degrees east refers to the longitude of the place. latitude-longitude are the coordinates of the grid system on Earth which helps in navigation.

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Which biome would you not find trees or deep rooted plants in ?

Dafna1 [17]

The tundra because it's growing season is too short.

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3) A 900N mountain climber scales a 100m cliff. How much work is done by the mountain climber?

Y_Kistochka [10]

Answer:

90 kJ

Explanation:

Work = force × distance

W = (900 N) (100 m)

W = 90,000 J

W = 90 kJ

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3% of earth's water is?

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Is earth’s freshwater

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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

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon

Yuri [45]

consider the motion in y-direction

v₀ = initial velocity = 0 m/s

a = acceleration = g = - 9.8 m/s²

t = time

v = final velocity at any time "t"

velocity at any time is given as

v = v₀ + at

v = 0 + (- 9.8) t

v = (- 9.8) t

so at t = 1 , v = (- 9.8) (1) = - 9.8 m/s

at t = 2 , v = (- 9.8) (2) = - 19.6 m/s

at t = 3 , v = (- 9.8) (3) = - 29.4 m/s and so on

Y₀ = initial position where the marble was dropped from = 0 m

Y = final position at any time "t"

final position at any time "t" is given as

Y = Y₀ + v₀ t + (0.5) a t²

Y = 0 + (0) t + (0.5) (-9.8) t²

Y = - 4.9 t²

at t = 1 , Y = - 4.9 (1)² = - 4.9 m

at t = 2 , Y = - 4.9 (2)² = - 19.6 m

at t = 3 , Y = - 4.9 (3)² = - 44.1 m

and So on

6 0

3 years ago