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wolverine [178]
3 years ago
8

A 12m long gate that is 4 m wide is placed against a body of water at an angle of 60 degrees. The gate is being pulled by a rope

placed 3 m from the top of the gate at 60 kN. If the water is 9 m deep, what is the minimum weight of the gate that will cause it to rotate clockwise
Physics
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

Since the weight cannot take a negative value, it is concluded that the minimum weight required for the gate is zero, because the same pressure force will rotate the gate.

Explanation:

Given:

L = 12 m

F = 60 kN

w = 4 m

H = 9 m

The pressure force is:

F_{p} =\rho gAx=1000*9.8*\frac{9}{sin60} *4*\frac{9}{2} =1833.203kN

The center of pressure is:

h=\frac{\frac{(\frac{9}{sin60})^{3} *4 }{12}*sin^{2}60  }{\frac{9}{sin60}*4*\frac{9}{2}  } +\frac{9}{2} =6

If the moment is equal to zero,

F_{p} *(9-6)+W*\frac{L}{2} cos60=F*9\\1833.203*3+W*6*cos60=60*9\\W=-1653.2kN

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The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo
spin [16.1K]

Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

W=\vec F\ \vec r

It can be written in its scalar version as

W=F.d.cos\theta

Being F and d the magnitudes of the force and displacement, and \theta the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

W=N.d.cos90^o=0

The work is zero because the force and the displacement are perpendicular

The static friction force exists only when the object is being applied a force of a magnitude not large enough to produce movement, i.e. the object is at rest. If the object is moved, the friction force is still present, but it's called dynamic friction force, usually smaller than the static.

Since in this case, there is no displacement, d=0, and the work is

W=F_r(0)cos180^o=0

3 0
3 years ago
Removing an electron from a neutral atom will result in an atom that?
Alex73 [517]

Removing an electron from a neutral atom will result in an atom that is positive.

8 0
2 years ago
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Jenise is buying a car for $7,020. The TAVT rate is 9.1%.
djyliett [7]

Answer:

$7,658.82

Explanation:

<u>Sales Tax Calculations:</u>

Sales Tax Amount = Net Price x (Sales Tax Percentage / 100)

Total Price = Net Price + Sales Tax Amount

Net Price: $ 7,020.00

+Sales Tax (9.1%): $ 638.82

Total Price: $ 7,658.82

Therefore, the amount of tax that Jenise has to pay on her car is $7,658.82

8 0
3 years ago
You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear
Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

r_2 =15 \times 10^{-1.5}

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0
3 years ago
Can anyone plz answer this now and give me a right answer?
klemol [59]

Answer:

90 degree hope it help :))

5 0
2 years ago
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