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Contact [7]
4 years ago
14

A rigid dam is composed of material of SG = 5 . The dam height is 20 m . What is the minimum thickness b of the dam necessary to

prevent it from tipping about the point O when the water reaches the top of the dam considering unit width? Assume that the maximum hydrostatic pressure acts over the bottom of the dam.
Physics
1 answer:
Kay [80]4 years ago
3 0

Answer:

b = 5.164 m is the minimum thickness of the dam

Explanation:

Given

SG = 5

h = 20 m

b = ?

w = 1 m

γw = 9800 N/m³

We  can get the forces as follows

Fp =  γw*(h/2)*(h*w)

⇒ Fp = (9800 N/m³)*(20 m/2)*(20 m*1 m) = 1.96*10⁶N

W = (SG*γw)*(h*b*w)

⇒  W = (5*9800 N/m³)*(20 m*b*1 m) = (9.8*10⁵N/m)*b

Then, we apply

∑M₀ = 0  (counterclockwise)

- Fp*(h/3) + W*(b/2) = 0

⇒   - 1.96*10⁶N*(20 m/3) + (9.8*10⁵N/m)*b*(b/2) = 0

- 13.066*10⁶N-m + (4.9*10⁵N/m)*b² = 0

⇒  b = 5.164 m is the minimum thickness of the dam

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3 years ago
Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

\displaystyle \frac{d' }{d}=4

Thus:

d' = 4d

The distance is now 4d

3 0
3 years ago
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33××10âˆ
AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, F=3.33\times 10^{-21}\ N

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

F=k\dfrac{q^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}

q=1.21\times 10^{-16}\ C

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

n=\dfrac{q}{e}

n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0
4 years ago
Can an object have a low acceleration and high velocity at the same time . give example
LenaWriter [7]
Yes, acceleration only tells you how velocity is changing. It doesn't say anything about what velocity is at any given time.

For example, if you set your car to cruise control on the highway going 80 mph. That is a constant high velocity, yet the car has 0 acceleration.

The opposite is also true. After a red light turns green, you put foot on the gas to accelerate. However, your velocity is initially low even though it has high acceleration.
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3 years ago
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