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Gre4nikov [31]
1 year ago
6

Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com

e out of the holes if these have a diameter of 40mm? At what speed would the water come out if the diameter is 120 mm?
Physics
1 answer:
Lubov Fominskaja [6]1 year ago
7 0

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm   ==>   area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

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In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component
Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0
2 years ago
A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe
jenyasd209 [6]

answer;

The hole in the center of the washer will expand

explanation;

<em>A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? The hole in the center will remain the same size. Changes in the hole cannot be determined without know the composition of the metal. The hole in the center of the washer will expand. The hole in the center of the washer will contract.</em>

this is an example of area expansivity.

coefficient of area expansivity is change in area per area per degree rise in temperature

a=dA/(A*dt)

as the temperature rises , there will be volumetric and area expansivity on the body. volume also increases because of the intermolecular forces of attraction between the molecule is now getting apart.

7 0
2 years ago
It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging riv
zzz [600]

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

V_2^2=V_o^2-2*g*\Delta h

V_2^2=-2*10*(-31)

V_2=24.9m/s   This will be our initial speed for a balance of energy.

By conservation of energy:

m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0

Where

h is your height at its maximum elongation

h_o is the height of the bridge

l_o is the length of the unstretched bungee cord

800h+21*(79-h)^2-63200-24800.4=0

21h^2-2518h+43060.6=0 Solving for h:

h_1=20.66m  and h_2=99.24m  Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

6 0
2 years ago
Read 2 more answers
A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to
Maurinko [17]

Answer:

v = 0.8 m/s towards left

Explanation:

As we know that there is no external force on the system of two cart so total momentum of the system is conserved

so we will say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now plug in all data into the above equation

3(1) + 5(-2) = 3(-1) + 5 v

here we assumed that left direction of motion is negative while right direction is positive

so we can solve it for speed v now

3 - 10 = - 3 + 5 v

5 v = -4

v = -0.8 m/s

3 0
2 years ago
An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini
NNADVOKAT [17]

Answer:

 E_total = 5.8 10⁴ N /C

Explanation:

In this problem they ask to find the electric field at two points, the electric field is a vector magnitude, so we can find the field for each charged shoah and add them vectorally at the point of interest.

To find the electric field of a charged conductive sheet, we can use the Gauss law,

        Ф = E. d S = q_{int} / ε₀

Let us use as a Gaussian surface a small cylinder, with the base parallel to the sheet, the electric field between the sheet and the normal one next to the cylinder has 90º, so its scalar product is zero, the electric field between the sheet and the base has An Angle of 0º, therefore the scalar product is reduced to the algebraic product.

Let's look for the electric field for plate 1

The total flow is the same for each face, as there are two sides of the cylinder

       2E A = q_{int} /ε₀

For the internal load we use the concept of surface density

      σ = q_{int1} / A

      q_{int1} = σ₁ A

Let's replace

       2E A = σ₁ A /ε₀

        E₁ = σ₁ / 2ε₀

For the other plate we have a field with a similar expression, but of negative sign

       E₂ = -σ₂ / 2ε₀

The total field is,

        E_total = σ₁ / 2ε₀ + σ₂ / 2ε₀

       E_total = (σ₁ + σ₂) / 2ε₀

Let us apply this expression to our case, when placing a sheet without electric charge, a charge is induced for each sheet, the plate 1 that has a positive charge the electric field is protruding to the right and the plate 2 that has a negative charge creates a incoming field, to the right, as the two fields have the same address add

           The conductive sheet in the middle pate undergoes an induced load that is created by the other two plates, but because the conductive plate the charges are mobile and are replaced.

       E_total = (0.51 +0.52) 10⁻⁶ / 2 8.85 10⁻¹²

       E_total = 5.8 10⁴ N /C

Note that the field is independent of the distance between the plates

4 0
2 years ago
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