If you are good at activities that require agility, what are you able to do well?

A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second

Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0

4 years ago

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

PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS<br>

butalik [34]

Answer:

g=GM/R^2

Universal Gravutation Constant:

f=GM×m/R^2

Force can be also expressed as

f=m×g

so,

mg=GMxm/R^2

The m gets cancelled so

g=GM/R^2

8 0

3 years ago

A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C. How much water freezes onto the ice? The specific heat of

vladimir2022 [97]

For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice? is mathematically given as

x = 9.93 g

<h3>What is the amount of water that freezes onto the ice?</h3>

Where

Energy received = energy given out

Generally, the amount of water is mathematically given as

(53)(0.5)(30) = (80)(x)

Therefore

x = (49)(0.5)(16)/(80)

x = 9.93 g

In conclusion, the mass of water

x = 9.93 g

Read more about mass

brainly.com/question/15959704

4 0

2 years ago

While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000

Minchanka [31]

Answer:

a

The hiker (you ) is 200 m below his/her(your) starting point

b

The resultant displacement in the north east direction is

$a = 6562.0 \ m$

The resultant displacement in vertical direction (i.e the altitude change )

$b =6503.1 \ m$

Explanation:

From the question we are told that

The displacement in the morning is $S_{morning} = (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)$

The displacement in the afternoon is $S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)$

Generally the direction west is negative , the direction east is positive

the direction south is negative , the direction north is positive

resultant displacement is mathematically evaluated as

$(2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east$

$(4000\ m\ north) + (2500 \ m ,\ north) = 6500 \ m ,\ north$

$(100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m$

From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

$a = \sqrt{900^2 + 6500^2}$

=> $a = 6562.0 \ m$

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

$b = \sqrt{6500^2 +(-200)^2 }$

=> $b =6503.1 \ m$

5 0

3 years ago