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mr Goodwill [35]
3 years ago
8

When cartographers represent the three-dimensional earth in two dimensions, what is likely to occur?

Physics
2 answers:
Harlamova29_29 [7]3 years ago
7 0
Distortion is likely to occur
Sladkaya [172]3 years ago
6 0

distortion is most likely to occur.

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a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0
3 years ago
Evidence on why we don’t hear solar eruptions on Earth.
sukhopar [10]
Because sound waves don't travel through the vaccume of space. Hope this helped
8 0
2 years ago
Read 2 more answers
A 0.1 kg beach ball hit me moving forward at 4 m/s. After hitting me, it bounced back moving at the same speed
xenn [34]

0.4 N-s is the "impulse" acted on the "beach ball".

Option: C

Explanation:

Given that,

Mass of the "beach ball" is 0.1 kg.

The speed of the ball hits is 4 m/s.

We know that,

Whenever an object is collide with other object then an impulse is acted on object, this "impulse" causes "change in momentum".

Impulse acted on the beach ball is "mass" times "velocity".

Impulse = mass × velocity

Impulse = 0.1 × 4

Impulse = 0.4 kg m/s

Impulse = 0.4 N-s

Therefore, the "impulse" acted on the ball is 0.4 N-s.

7 0
3 years ago
Light in a vacuum travels at a constant speed of 3x10^8 m/s. If the moons average distance from the earth is 38776106 km how lon
Karo-lina-s [1.5K]

Answer: 258.3 s

Explanation:

The speed s is given by the following equation:

s=\frac{D}{t}

Where:

s=3(10)^{8} m/s is the speed of light in vacuum

D=2(38776106 km \frac{1000 m}{1 km})=7.75(10)^{10} m is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.

t is the time it takes to the beam of light to travel the mentioned distance

Isolating t and solving with the given information:

t=\frac{D}{s}

t=\frac{7.75(10)^{10} m}{3(10)^{8} m/s}

Finally:

t=258.3 s \approx 258 s

5 0
3 years ago
Question 2 please help
Sergeeva-Olga [200]

Answer:

D

Explanation:

3 0
3 years ago
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