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Kruka [31]
3 years ago
8

A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet A. Knowing that the truck starts from rest

and that the horizontal force between the tires and the ground is 700 lb, determine the velocity of the boulder after the truck has moved forward 6 ft.
Engineering
1 answer:
artcher [175]3 years ago
4 0

Answer:

Explanation:

Total weight being moved = 5000+1000+200

= 6200 lb .

Force applied = 700 lb

= 700 x 32 = 22400 poundal .

acceleration (a) = 22400 / 6200

= 3.613 ft /s²

To know velocity after 6 ft we apply the formula

v² = u² + 2as

v² = 0 + 2 x 3.613 x 6

43.356

v = 6.58 ft/s

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White lane lines separate lanes of traffic moving in the same direction. (UK)

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A eutectic alloy is which of the following (two correct answers): (a) the composition in an alloy system at which the two elemen
NeTakaya

Answer:

Option (c) and option (d)

Explanation:

Eutectic system is one in which a solid and homogeneous mixture of two or more substances resulting in the formation of super lattice is formed which can melt or solidify at a temperature lower than the melting point of any individual metal.

Eutectic alloys are those which have its components mixed in a specific ratio.

It is the composition in an alloy system for which both the  liquidus and solidus temperatures are equal.

Eutectic alloys have the composition in which the melting point of the metal is lower than the other alloy composition.

3 0
3 years ago
2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks
Amiraneli [1.4K]

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

4 0
3 years ago
Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai
Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

5 0
3 years ago
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s
monitta

Answer:

a. \dfrac{D_{1}}{ D_{2}}  =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n} which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

v_1 = \dfrac{4}{3} \times  \pi \times r^3

\therefore v_1 = \dfrac{4}{3} \times  \pi \times  \left (\dfrac{10}{2}  \right )^3 = 523.6 \ m^3

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} = \left (\dfrac{T_{1}}{T_{2}}   \right )^{\dfrac{n}{n-1}}

We have;

\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_2}{2}  \right )^3}{\dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_1}{2}  \right )^3}   \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2  \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

log  \left (\dfrac{D_{1}}{ D_{2}}\right )  =  -3\times n \times log\left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

\left (\dfrac{628.32 }{523.6}   \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{1}{4}}

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}

p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} } =  \dfrac{100\times 10^3}{ \left (1.2) \right  ^{-\dfrac{1}{3} } }

p₂ =  100000/0.941 = 106.265 kPa

W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3  \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0
3 years ago
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