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Triss [41]
3 years ago
13

Julio blows air across his hot bowl of soup. The tiny ripples he creates are similar to _____.

Physics
2 answers:
Allushta [10]3 years ago
8 0

The tiny ripples on the soup are not only similar to wind-generated
waves ... they ARE wind-generated waves.  This is a big part of the
reason why they bear such an uncanny resemblance.

mojhsa [17]3 years ago
6 0

Answer: Option (c) is the correct answer.

Explanation:

Wind-generated waves means formation of waves at the surface of water or fluid.  

For example, when Julio blows air across his hot bowl of soup then tiny ripples he creates are like wind generated waves. As they are moving in circular motion.

Thus, we can conclude that the tiny ripples Julio creates are similar to wind-generated waves.

You might be interested in
A student attaches a rope to his
erastova [34]

The choices are:

a. Normal Force

b. Gravity Force

c. Applied Force

d. Friction Force

e. Tension Force

​f. Air Resistance Force

Answer:

The answer is letter e, Tension Force.

Explanation:

Force refers to the "push" and "pull" of an object, provided that the object has mass. This results to acceleration or a change in velocity. There are many types of forces such as <em>Normal Force, Gravity Force, Applied Force, Friction Force, Tension Force and Air Resistance Force.</em>

The situation above is an example of a "tension force." This is considered the force that is being applied to an object by strings or ropes. This is a type force that allows the body to be pulled and not pushed, since ropes are not capable of it. In the situation above, the tension force of the rope is acting on the bag and this allows the bag to be pulled.

Thus, this explains the answer.

6 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
3 years ago
A point moves on the x-axis in such a way that its velocity at time t (t &gt; 0) is given by v=ln t/t . At what value of t does
Olenka [21]

Answer:

Explanation:

Given

Velocity of point is given by v=\frac{\ln t}{t}

To get maximum or minimum velocity differentiate v w.r.t t

\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\frac{1}{t}\times t-1\times \ln(t)}{t^2}

so 1-\ln (t) should be equal to zero

\ln (t)=1

t=e

i.e. t=2.718\ s

5 0
3 years ago
Pls help me quickly ......​
marishachu [46]

Answer:

the last one: weight force

3 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But,  the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0
2 years ago
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