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sineoko [7]
3 years ago
11

A particle moves along the x-axis so that its velocity at anytime is t greater than or equal to 0 is given by v(t)=1-sin(2pi t)

Physics
1 answer:
UkoKoshka [18]3 years ago
5 0
A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)

b) Any value of t can be plugged in as long as it is greater than or equal to 0. 

c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π
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If a wave has amplitude of 2 meters, a wavelength of 2 meters, and a frequency of 10 Hz, and a period of 1 second, then at what
Serhud [2]

Answer:

20 m/s

Explanation:

The speed of a wave is given by:

v=\lambda f

where

\lambda is the wavelength

f is the frequency

v is the speed

For the wave in this problem,

f = 10 Hz is the frequency

\lambda=2 m is the wavelength

So the speed is

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6 0
3 years ago
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
3 years ago
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s
notka56 [123]

Answer:

     P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

           I = P / A

Since sound is distributed in all directions spherical shape, the area of ​​a sphere is

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let's clear the power and replace

         P = I A

         P = I (4π r²)

let's calculate

         P = 5.00 (4π 2²)

         P = 251, 3 W

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Answer:

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