Mathematics College

For a super soaker water gun, a pump handle is moved back and forth to build up pressure in the water reservoir. The water is released by pulling a trigger and shooting the water a significant distance. Assuming that you can create an absolute pressure of 8 atm in the reservoir:
a) What is the velocity at which the water leaves the gun?
b) If the water exits the gun through a hole with a radius of 1-mm, what is the volume rate of flow in m3/s?
c) If the water gun is fired horizontally and held 1.2 meters above the ground, where does the water hit the ground? Pressure 8 cm water

Answers

Answer 1

Answer:

a) The velocity at which the water leaves the gun = 37.66 m/s

b) The volume rate of flow = (1.183 × 10⁻⁴) m³/s

c) The water hits the ground 18.64 m from the point where the water gun was shot.

Step-by-step explanation:

a) Using Bernoulli's equation, an equation that is based on the conservation of energy.

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

The two levels we are considering is just inside the water reservoir and just outside it.

ρgh is an extension of potential energy and since the two levels are at the same height,

ρgh₁ = ρgh₂

Bernoulli's equation becomes

P₁ + ½ρv₁² = P₂ + ½ρv₂²

P₁ = Pressure inside the water reservoir = 8 atm = 8 × 101325 = 810,600 Pa

ρ = density of water = 1000 kg/m³

v₁ = velocity iof f water in the reservoir = 0 m/s

P₂ = Pressure outside the water reservoir = atmospheric pressure = 1 atm = 1 × 101325 = 101,325 Pa

v₂ = velocity outside the reservoir = ?

810,600 + 0 = 101,325 + 0.5×1000×v₂²

500v₂² = 810,600 - 101,325 = 709,275

v₂² = (709,275/500) = 1,418.55

v₂ = √(1418.55) = 37.66 m/s

b) Volumetric flowrate is given as

Q = Av

A = Cross sectional Area of the channel of flow = πr² = π×(0.001)² = 0.0000031416 m²

v = velocity = 37.66 m/s

Q = 0.0000031416 × 37.66 = 0.0001183123 m³/s = (1.183 × 10⁻⁴) m³/s

c) If the height of gun above the ground is 1.2 m. Where does the water hit the ground?

The range of trajectory motion is given as

R = vT

v = horizontal component of the velocity = 37.66 m/s

T = time of flight = ?

But time of flight is given as

T = √(2H/g) (Since the initial vertical component of the velocity = 0 m/s

H = 1.2 m

g = acceleration due to gravity = 9.8 m/s²

T = √(2×1.2/9.8) = 0.495 s

Range = vT = 37.66 × 0.495 = 18.64 m

Hope this Helps!!!

Related Questions

I NEED HELP WITH THIS PLEASE HELP ME

Answers

Answer:

156 minutes

Step-by-step explanation:

So we need to create an equation to represent how Frank's phone company bills him

I will denote "y" as the total for his billI will denote "x" as the number of minutes Frank uses

So the phone company charges an $8 monthly fee, so this value remains constant and will be our "y-intercept"

They then charge $0.06 for every minute he talks, this will be our "slope"

Combining everything into an equation, we have: y = 0.06x + 8

Now since we were given Franks phone bill total and want to figure out how many minutes he used, we just need to solve the equation for x and plug in our known y value

y = 0.06x + 8 → y - 8 = 0.06x → [tex]x=\frac{y-8}{0.06}[/tex] Then plugging in our y value we get [tex]x=\frac{17.36-8}{0.06}=\frac{9.36}{0.06}= 156[/tex]

Frank used up a total of 156 minutes

(Bonus) A rectangular box has its edges changing length as time passes. At a par-ticular instant, the sides have lengthsa= 150 feet,b= 80 feet, andc= 50 feet.At that instant,ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec. Determine if the volume of the box is increasing, decreasing,or not changing at all, at that instant.

Answers

Answer:

the volume of the box is increasing

dV = +310,000 ft^3/s

Step-by-step explanation:

Volume of a rectangular box with side a,b and c can be expressed as;

V = abc

The change in volume dV can be expressed as;

dV = d(abc)/da + d(abc)/db + d(abc)/dc

dV = bc.da + ac.db + ab.dc ......1

Given:

a= 150 feet,

b= 80 feet, and

c= 50 feet

ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec

da = +100 feet/s

db = -20 feet/s

dc = +5 feet/s

Substituting the values into the equation 1;

dV = (80×50×+100) + (150×50×-20) + (150×80×+5)

dV = +400000 - 150000 + 60000 ft^3/s

dV = +310,000 ft^3/s

Since dV is positive, the volume of the box is increasing at that instant.

Two negative integers are 8 units apart on the number line and have a product of 308. Which equation could be used to determine x, the smaller negative integer? A: x^2 + 8x – 308 = 0 B: x^2 – 8x + 308 = 0 C: x^2 + 8x + 308 = 0 D: x^2 − 8x − 308 = 0

Answers

Answer:

A

Step-by-step explanation:

The smaller negative integer is x.

The larger one is x+8, since they are 8 units apart.

The equation would be:

x*(x+8)=308

Let's simplify it by distributing.

x^2+8x=308

Subtract 308 from both sides.

x^2+8x-308=0

Therefore, the answer would be A.

A bookstore charges $4 for shipping, no matter how many books you buy. Irena makes a graph showing the shipping cost for I to 5 books. She claims that the points she graphed lie on a line. Does her statement make sense? Explain

Answers

Answer:

Yes

Step-by-step explanation:

1 book = $4

2 books = 2*$4

3 books = 3*$4

4 books = 4*$4

5 books = 5*$4

This can be shown as:  y=4x

y=ax+b is linear function, Irena is right

Evaluate 16x^0 if x= -3

Answers

Answer:

16

Step-by-step explanation:

[tex]16x^0= \\\\16(-3)^0= \\\\16(1)= \\\\16[/tex]

Hope this helps!

x = -3

[tex]A = 16.(-3)^{0} \\ x^{0} = 1\\A = 16.1 \\A = 16[/tex]

Remember that [tex]x^{0} = 1[/tex] ∀ [tex]x[/tex]

Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
lim √9 + h - 3 / h
h-->0

Answers

Answer:

a = 0f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]limit of the function is 1/6

Step-by-step explanation:

The general form representing limit of a function is expressed as shown below;

[tex]\lim_{h \to a} f(h)[/tex] where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question [tex]\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}[/tex], it can be seen that a = 0 and f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]

Taking the limit of the function

[tex]\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)[/tex]

Applying l'hopital rule

[tex]\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6[/tex]

Other Questions
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