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3 years ago

2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks

Engineering

1 answer:

Amiraneli [1.4K]3 years ago

4 0

Answer:

100N
25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

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A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

3 years ago

Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus

Alina [70]

Answer:

Explanation:

find attached the solution to the question

4 0

3 years ago

Is a jeep cherokee faster than a bmw 325i

soldier1979 [14.2K]

Answer:

Yes

Explanation:

8 0

3 years ago

Read 2 more answers

A water reservoir contains 108 metric tons of water at an average elevation of 84 m. The maximum amount of electric energy that

zavuch27 [327]

Answer:

24.72 kwh

Explanation:

Electric energy=potential energy=mgz where m is mass, g is acceleration due to gravity and z is the elevation.

Substituting the given values while taking g as 9.81 and dividing by 3600 to convert to per hour we obtain

PE=(108*9.81*84)/3600=24.72 kWh

8 0

4 years ago

Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given

Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = $\frac{1}{6}*334.8 = 55.8 ft^3$

Volume of sand = $\frac{2}{6}*334.8 = 111.6 ft^3$

Volume of gravel = $\frac{3}{6}*334.8 = 167.4 ft^3$

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0

3 years ago