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Serga [27]
2 years ago
10

The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag

e depth of the water in kilometers. What is the approximate depth of water for a tsunami traveling at 200 kilometers per hour?.
Physics
1 answer:
Goshia [24]2 years ago
7 0

0.32 km

How to reach the answer:

The formula below provides the tsunami's speed in response to the query.

s = 356√d

when the tsunami's speed is 200 km/h, the equation's "s" is changed to "200," and the equation of speed is then expressed as follows.

200 = 356√d

The equation can be changed to compute the depth of the water as,

√d = 200÷356

√d = 0.562

Now on squaring both sides:

(√d)² = (0.562)²

d = (0.562)² = 0.316 or 0.32

Therefore the approximate depth of water for a tsunami travelling at 200km/h is 0.32 km

Know more about speed here:

brainly.com/question/19930939

#SPJ4

​

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500

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Celine has raised the temperature of water to 100°C in order to cook pasta for her lasagna. What physical property is Celine dem
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3 years ago
Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
zhenek [66]

Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0
3 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?
yKpoI14uk [10]

Answer:

<h3>The answer is 5.4 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{135}{25}   =  \frac{27}{5} \\

We have the final answer as

<h3>5.4 kg</h3>

Hope this helps you

4 0
3 years ago
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