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Contact [7]
3 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]3 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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Which of the following is a major way in which oceans contribute to weather systems?
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Explanation:

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If a snowboarder’s initial speed is 4 m/s and comes to rest when making it to the upper level. With a slightly greater initial s
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(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

<h3>Conservation of mechanical energy</h3>

The effect of height  and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;

ΔK.E = ΔP.E

¹/₂m(v²- u²) = mg(hi - hf)

¹/₂(v²- u²) = g(0 - hf)

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v² = u² - 2ghf

where;

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  • u is the initial velocity
  • hf is final height
  • g is acceleration due to gravity

when u² = 2gh, then v² = 0,

when gravity reduces, u² > 2gh, and v² > 0

Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

<h3>Final speed</h3>

v² = u² - 2ghf

where;

  • u is the initial speed = 5 m/s
  • g is acceleration due to gravity and its less than 9.8 m/s²
  • v is final speed
  • hf is equal height

Since g on Epislon is less than 9.8 m/s² of Earth;

5² - 2ghf > 3 m/s

Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

Learn more about conservation of mechanical energy here: brainly.com/question/6852965

5 0
2 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

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0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

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2.

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Part a)

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Part b)

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Explanation:

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As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

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j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

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i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
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