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3 years ago

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A mover loads a 100 kg box into the back of a moving truck by pushing it up a ramp. The ramp is 4 m long and the back of the tru

ck is 1.5 m high. Calculate the force required to push the box up the ramp in the absence of friction.

1 answer:

elixir [45]3 years ago

6 0

Answer:

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Its not b...................

vladimir1956 [14]

sir what's the question you have all you wrote is it's not b

4 0

3 years ago

Agri-Chem's contract with Enerco specified a maximum of 90,000 cu.ft. X 103 per day for its complexes. However, curtailments are

zvonat [6]

Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

Agri chem’s current natural gas usage

= (1200 × 8 + 540 × 10 + 490 × 12 + …)

= 85,680,000 cu. ft. per day

When, the curtailment is 20%, availability is

= 0.8 X 85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

= 0.6 X 85,680

= 51,408,000 cu. ft. per day

Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408

7 0

3 years ago

A 4kg bird is flying with a velocity of 4 m/s. What is its kinetic energy?

Yanka [14]

Kinetic energy = 32 J

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2 years ago

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/

Hunter-Best [27]

Answer:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

$n_{T} = n_{1} + n_{2}$

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$

$n_{T} = 31.68\,rev$

4 0

3 years ago

A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the

hodyreva [135]

Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?

The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m

6 0

2 years ago