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Fofino [41]
2 years ago
9

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

When the pail is at the bottom of the circle, what is the magnitude of the force exerted by the water on the bottom of the pail?
Physics
1 answer:
Alisiya [41]2 years ago
7 0

Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path  so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of  vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r  

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

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mars1129 [50]

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2 years ago
A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a
Yanka [14]

Answer:

Diameter of the cylinder will be d=2.998\times 10^4m

Explanation:

We have given young's modulus of steel E=207GPa=207\times 10^9Pa  

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Length of rod l=500mm

Load F = 11100 KN

Strain is given by strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}

We know that young's modulus E=\frac{stress}{strain}

So 207\times 10^9=\frac{stress}{7.6\times 10^{-4}}

stress=1573.2\times 10^{-5}N/m^2

We know that stress =\frac{force}{artea }

So 1573.2\times 10^{-5}=\frac{11100\times 1000}{area}

area=7.055\times 10^{8}m^2

So \frac{\pi }{4}d^2=7.055\times 10^{8}

d=2.998\times 10^4m          

6 0
2 years ago
a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the
irinina [24]
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
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m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


7 0
2 years ago
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

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The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

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       v is the flow velocity

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substituting values

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Then

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Answer:

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a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}

8 0
2 years ago
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