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1 year ago

You are pulling a sled with a constant velocity using a rope held horizontally in

Physics

1 answer:

Mashcka [7]1 year ago

5 0

Answer:

Explanation:

if you are pulling in the positive direction and Upward is also positive.

F = -0.96i + 0.58(9.8)j

F = -0.96i + 5.7j

F = √(-0.96² + 5.7²) = 5.8 N

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2 years ago

A container of soy milk has the instructions on its label to “shake well before opening ” The soy milk is most likely a ?

Elena-2011 [213]

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D. Suspension

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2 years ago

Which of the following is right?

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2 years ago

A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago

] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0

2 years ago