Which is true about a surface wave? ANSWER FAST

A caterpillar tries to climb straight up a wall a meter high, but for every 2 cm up it climbs, it slides down 1 cm. Eventually,

tiny-mole [99]

<u>Answer:</u>

Total displacement traveled = 298

<u>Explanation:</u>

According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).

So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.

Then after a little it can travel up another cm to reach the top.

Therefore, the total displacement traveled = 297 + 1 = 298 cm

5 0

3 years ago

Read 2 more answers

Peter notices that his eight-year-old daughter frequently mentions the unusual habits of a new boy, who is from a different cult

pickupchik [31]

The answer should be C. Encourage his daughter to make friends with the new boy.

Hope this helps and have a happy new year

6 0

3 years ago

A thin plate 6 ft long and 3 ft wide is submerged and held stationary in a stream of water (T = 60°F) that has a velocity of 17

VashaNatasha [74]

A) In the case of the Boundary Thickness Layer we use the given formula,

$\delta = \frac{4.91x}{\sqrt{Re}}$

We know as well that,

Re = Número de Reynolds = $\frac{U*x}{\upsilon}$

Where,

U = velocity

$\upsilon$ = kinematic viscosity

For water, kinematic viscosity, $\upsilon = 1.21*10^{-5} ft^2 /s$

So, $500,000 = \frac{ 17x}{(1.21*10^{-5})}$

$x = 0.355 ft$

$d = \frac{4.91*0.355}{\sqrt {500000}}$

$d = 0.002465 ft = 0.029in$

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, $x = 0.355 ft$

C. Wall shear stress,

$\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }$

For water, dynamic viscosity, $\nu$ = 2.344*10^-5 lbf-s/ft^2

$\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}$

$\tau = 0.5605 lbf/ft^2$

4 0

3 years ago

a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height

ki77a [65]

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′ and mass ′

M′ is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

6 0

3 years ago

Help me with this speed thingy I don't have no idea

Aneli [31]

He goes 24 miles. hope this helps!

7 0

3 years ago

Read 2 more answers