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UkoKoshka [18]
3 years ago
14

The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru

e is 400 mm2 . The mold cavity has a volume of 0.001 m3 . Determine (a) the velocity of the molten metal flowing through the base of the downsprue, (b) the volume rate of flow, and (c) the time required to fill the mold cavity.
Physics
1 answer:
adoni [48]3 years ago
5 0

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

(c) The time required to fill the mold is 1.35 s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

<u>V = 1.85 m/s</u>

<u></u>

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

<u>Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s</u>

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

<u>t = 1.35 s</u>

<u></u>

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Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

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Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²

3 0
3 years ago
A ball is moving with a velocity of 0.5 m/s its velocity is decreasing at the rate of 0.05 m/s² what is its velocity in 5 second
Law Incorporation [45]

Answer:

v after 5s = 0.25 m/s, it took 10s to stop, it has traveled 2.5m before stopping

Explanation

We can use the equation of motion with constant acceleration

Given: v0= 0.5 m/s a= -0.05 m/s²

v(5s) = v0 + a×t = 0.25 m/s

Stop => v=0 => v0 + a×t = 0 => t=10s

Distance at t=10s ⇒ x(10) = 0.5×10 + 0.5x(-0.05)x10² = 2.5m

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3 years ago
Easy question. Answer should include one line reasoning
KATRIN_1 [288]
C.figure 3 is the answer had the same and got is right
7 0
3 years ago
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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

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At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
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Answer:

Explanation:

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