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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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A 6.85-m radius air balloon loaded with passengers and ballast is floating at a fixed altitude. Determine how much weight (balla
melisa1 [442]

Answer: 120 kg

Explanation:

Given

Radius of balloon, r = 6.85 m

Distance moved by the balloon, d = 114 m

Time spent in moving, t = 17 s

Density of air, ρ = 1.2 kg/m³

Volume of the balloon = 4/3πr³

Volume = 4/3 * 3.142 * 6.85³

Volume = 4/3 * 3.142 * 321.42

Volume = 4/3 * 1009.90

Volume = 1346.20 m³

Density = mass / volume ->

Mass = Density * volume

Mass = 1.2 * 1346.2

Mass = 1615.44 kg

Velocity = distance / time

Velocity = 114 / 17

Velocity = 6.71 m/s

If it starts from rest, 0 m/s, then the final velocity is 13.4 m/s

acceleration = velocity / time

acceleration = 13.4 / 17 m/s²

The mass dropped from the balloon decreases Mb and increases buoyancy

F = ma

mg = (Mb - m) * a

9.8 * m = (1615.44 - m) * 13.4/17

9.8m * 17/13.4 = 1615.44 - m

12.43m = 1615.44 - m

12.43m + m = 1615.44

13.43m = 1615.44

m = 1615.44 / 13.43

m = 120.29 kg

5 0
3 years ago
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
N76 [4]

Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

7 0
3 years ago
Jeff is pushing a rock up a hill with a force of 85 N. If Chris comes to help and Chris can push with a force of 60 N, what is t
Oksanka [162]

Answer:

F = 145 N

Explanation:

Given that,

Force applied by Jeff is 85 N and the force applied by Chris is 60 N. Both forces are acting in the same direction to push a rock up a hill. It means that net force is equal to the sum of these two forces.

So,

Net force on the rock = 85 N + 60 N

= 145 N

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7 0
3 years ago
A spring is used to launch a coffee mug. The 20cm long spring can be compressed by a maximum of 8cm. The mug has a mass of 500 g
Otrada [13]

Answer:

7664.06249 N/m

Explanation:

m = Mass of mug = 0.5 kg

x = Compression of spring = 0.08 m

h = Height of fall = 5 m

g = Acceleration due to gravity = 9.81 m/s²

The kinetic energy of the spring and potential energy of the fall are conserved

\frac{1}{2}kx^2=mgh\\\Rightarrow k=\frac{2mgh}{x^2}\\\Rightarrow k=\frac{2\times 0.5\times 9.81\times 5}{0.08^2}\\\Rightarrow k=7664.06249\ N/m

The spring constant of the spring is 7664.06249 N/m

8 0
3 years ago
if escape velocity and orbital velocity of a satellite for Orbit close to the Earth's surface then these are related by​
n200080 [17]

Answer:

The ratio of the escape velocity to the orbital velocity is \sqrt 2.

Explanation:

The minimum velocity given to an object so that it escapes from the earth's gravitational pull is called escape velocity.

The formula of the escape velocity is

v=\sqrt\frac{2GM}{R}

The velocity of an object in the orbit around the earth is called orbital velocity.

The formula of the orbital velocity is

v=\sqrt\frac{GM}{R}

The ratio of the escape velocity to the orbital velocity is

\frac{v_e}{v_o}=\sqrt 2

3 0
3 years ago
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