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SCORPION-xisa [38]
3 years ago
11

For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time

iv) Frequency v) On a diagram show the positions of max and min values for Acceleration and Velocity and show using the equations why this is the case.

Engineering
1 answer:
max2010maxim [7]3 years ago
4 0

Answer with Explanation:

The general equation of simple harmonic motion is

x(t)=Asin(\omega t+\phi)

where,

A is the amplitude of motion

\omega is the angular frequency of the motion

\phi is known as initial phase

part 1)

Now by definition of velocity we have

v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )

part 2)

Now by definition of acceleration we have

a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )

part 3)

The angular frequency is related to Time period 'T' asT =\frac{2\pi }{\omega }

where

\omega is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is A\omega ^{2} and A\omega respectively.

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The water requirement for Class H cement is 38% (i.e.,water (%) by weight of cement),whereas the water requirement for barite is
Vikentia [17]

Answer:

weight of barite = 398.4355 kg

Explanation:

Solution:- The values given in the question are as follows:

water requirement for H class cement = 38% by weight of cement

water requirement for barite = 2.4 gal / 100 lbm

H class cement slurry = 15.7 lbm/gal

one sack of cement = 50 kg or 110.231 lbm

one sack of cement require water = (38/100)*110.231

one sack of cement require water = 41.8877 gal

water required 100 lbm barite = 2.4 gal

or water required barite = 2.4% by weight of barite

H class cement slurry = (weight of cement + weight of barite)/total weight of water

15.7 =(110.231 + weight of barite)/(water required one sack of cement + 2.4%*weight of barite)

15.7 = 110.231 + (weight of barite)/(41.887 + 0.024*weight of barite)

15.7*41.8877 + 15.7*0.024*weight of barite = 110.231 + weight of barite

657.6368 + 0.3768*weight of barite = 110.231 + weight of barite

547.4058 =0.6232*weight of barite

weight of barite = 878.379 lbm or 878.4 lbm

weight of barite = 398.4355 kg

7 0
3 years ago
You start your car and begin to pull out of a parking space. After leaving the space, You notice that the brake light on your in
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If your brake lights are staying on, it will drain your battery.The most likely cause for the brake lights staying on is the brake light switch sticking closed. You can disconnect the plug from the brake light switch at the brake light pedal to see if the lights turn off.

Hope it helped!
7 0
3 years ago
Hỗ trợ mình với được không các bạn
Leya [2.2K]

Answer:

Explanation:

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3 years ago
Explicar el funcionamiento de un multímetro analógico.
Whitepunk [10]

Answer:

Un multímetro analógico funciona como un medidor de bobina móvil de imán permanente (PMMC) para tomar mediciones eléctricas

Explanation:

El multímetro analógico es un medidor o galvanómetro D'Arsonval que funciona según el principio de los medidores de bobina móvil de imán permanente (PMMC)

Un multímetro analógico está formado por un puntero de aguja unido a una bobina móvil colocada entre el polo norte y sur de un imán permanente dispuesto de tal manera que, cuando una corriente eléctrica fluye a través de la bobina, genera una fuerza de campo magnético que interactúa con el imán fuerza de campo de los imanes permanentes que hace que la bobina se mueva junto con el puntero de la aguja sobre un dial graduado

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4 0
3 years ago
A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i
Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature T_{1} = 300 F  = 422.04 K

Duct temperature T_{2} = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant \sigma = 5.67 × 10^{-8} \frac{W}{m^{2} K^{4}  }

Heat transfer  per foot length is given by

Q = ∈ \sigma A ( T_{1}^{4} - T_{2} ^{4} ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × 10^{-8} × 3.14 × 0.0508 × 1 × ( 422.04^{4} - 294.26^{4} )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0
3 years ago
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