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Katen [24]
3 years ago
12

Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

Physics
1 answer:
erica [24]3 years ago
5 0
Wave D has the same frequency
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Which can a drop of liquid mercury can be described as?
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(4%) Problem 9: A mass is connected to a spring and is allowed to move horizontally. The mass is at a position L when the spring
skad [1K]

Answer:

a) Acceleration is zero , c)   Speed ​​is cero

Explanation:

a) the equation that governs the simple harmonic motion is

         x = A cos (wt +φφ)

Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition

Body acceleration is

         a = d²x / dt²

Let's look for the derivatives

         dx / dt = - A w sin (wt + φ)

         a = d²x / dt² = - A w² cos (wt + φ)

In the instant when it is not stretched x = 0

As the spring is released at maximum elongation, φ = 0

            0 = A cos wt

            Cos wt = 0         wt = π / 2

Acceleration is valid for this angle

           a = -A w² cos π/2 = 0

Acceleration is zero

b)

c) When the spring is compressed x = A

Speed ​​is

             v = dx / dt

             v = - A w sin wt

We look for time

            A = A cos wt

            cos wt = 1         wt = 0, π

For this time the speedy vouchers

            v = -A w sin 0 = 0

Speed ​​is cero

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3 years ago
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A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of
Alexandra [31]

Answer:

The maximum no. of electrons- 2.25\times 10^{22}

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

I = \frac{Q}{t}

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

Q = 1.0\times 3600 = 3600\ C

Maximum number of electrons, n is given by:

n = \frac{Q}{e}

where

e = charge on an electron = 1.6\times 10^{- 19}\ C

Thus

n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}

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