1. What is wave motion

1 answer:

8 0

A wave is basically propagation of disturbances—that is, deviations from a state of rest or equilibrium—from place to place in a regular and organized way. Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties.

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scrieti in caiete substantivele ,adjectivele sau verbele apartinand familiei lexicale a urmatoarelor cuvinte din text : inima,ar

ser-zykov [4K]

Yes the answer is yes, all the way yes

5 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

(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas

aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = $\frac{k*q}{r}$