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Ugo [173]
3 years ago
15

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

of water is 62.4 lb/ft3).
Physics
1 answer:
blsea [12.9K]3 years ago
6 0

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

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An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
gayaneshka [121]

Answer:

Explanation:

Solution is in the picture attached

8 0
3 years ago
A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the
ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

a_{B} =a_{D}  +(a_{B/D} )_{t}

2.5j=1.5j  +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

2\alpha =1\\\alpha =0.5rad/s^{2}CW

b) The acceleration of point A is:

a_{A} =a_{D}  +(a_{A/D} )_{t}

(aA/D)t = ADαj

a_{A} =a_{D}  +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}

The acceleration of point E is:

(aE/D)t = -EDαj

a_{E} =a_{D}  -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}

7 0
3 years ago
PLEASE HELP LOTTA POINTS
sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question.  There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems.   That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick.  The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.  

When the question talks about hanging a weight, it tells the <em>centimeter mark on the stick</em> where the weight is tied.  To solve the problem, we have to first figure out <em>how far that is from the pivot</em>, then calculate how far from the pivot to put the weight on the other side, and finally <u><em>what centimeter mark that is</em></u> on the stick.      

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =  

#1).  Only one moment on the left side.  

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm)  =  9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot.  But that's not the answer to the problem.  They want to know what mark on the stick to hang it from.  The pivot is at the 50cm mark.  The 500gm has to hang 9.6 cm to the right of the pivot.  That's the <em>59.6 cm</em> mark on the stick.

= = = = =

#2).  There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight:  120gm, hanging from the 25cm mark.

That's 25cm from the pivot.  Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight:  20gm, hanging from the 10cm mark;

That's 40cm from the pivot.  Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick.  You have to hang the 500gm from 7.6cm to the right of that.  The mark at that spot on the stick is                (50cm + 7.6cm) = <em>57.6 cm </em>.

3 0
3 years ago
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when a wave crosses a surface at an angle from one medium into another, why does it change directions aS IT MOVES ACROSS THE BOU
puteri [66]
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Answer:

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Explanation:

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