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Natalka [10]
3 years ago
7

A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerati

ng it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.

Physics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

915 N is the anchoring force needed to hold the elbow in place.

Explanation:

Momentum equation:

∑

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Difference between compound and element?
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Explanation:

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1 year ago
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A 300 kg piano needs to be moved to the other side of the room. The maximum static frictional force is equal to 90 N and the kin
riadik2000 [5.3K]

Answer:

a = 0.1 m/s²

Explanation:

  • If the maximum static frictional force is 90 N, this means that any applied force that will overcome this force, will cause the piano to slide, so kinetic frictional force applies.
  • Under these conditions, the net force in the horizontal direction is just the difference between the applied force (which is larger that the static friction force) and the kinetic frictional force, as follows:

       F_{net}  = F_{app} -F_{frk}  = 100 N - 70 N = 30 N (1)

  • By the same token, according Newton's 2nd Law, this force is just equal to the product of the mass of the piano, times the acceleration, as follows:

       F_{net} = m* a = 300 Kg * a = 30 N (2)

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2 years ago
A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.
Anna [14]
Refer to the figure shown below.

g =  9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
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0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
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R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

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