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ASHA 777 [7]
3 years ago
8

35 miles North 6 feet down 136 MB 85 km SE

Physics
2 answers:
lys-0071 [83]3 years ago
8 0

Answer:

The answer is 136 MB I hope that this helps

Explanation:

pantera1 [17]3 years ago
4 0
The answer is 133 Se
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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 $100 cart is moving down an inclined plane at 1.30 m/s. The cart collides with a stationary object and rebounds with a speed o
Rasek [7]

Answer:

0.65  kg*m/s and 0.165 kg*m/s

Explanation:

Step one:

given data

mass m= 0.5kg

initial velolcity u=1.3m/s

final velocity v= 0.97m/s

Required

The change in momentum

Step two:

We know that the expression for impulse is given as

Ft= mv

Ft= 0.5*1.3

Ft= 0.65  kg*m/s

The expression for the change in momentum is given as

P= mΔv

substitute

Pt= 0.5*(1.3-0.97)

Pt= 0.5*0.33

Pt=0.165 kg*m/s

7 0
3 years ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
What is the answer to this definition ?
Paul [167]

Newton's 2nd law of motion is:    <em>Force</em> = (mass) x (acceleration) .

Force is the only way to change an object's velocity.


6 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area= \sigma = 1.99*10^{-7} C/m^2

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates isE = \frac{\sigma}{\epsilon}

force acting on charge is F = q E

Work done by charge q id\Delta X =\frac{ q\sigma \Delta x}{\epsilon}

this work done is converted into kinectic enerrgy

\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}

solving for v

v = \sqrt{\frac{2q\Delta x}{\epsilon m}

\epsilon = 8.85*10^{-12} Nm2/C2

v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}

v = 1.15*10^{7} m/s

8 0
3 years ago
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