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2 years ago

Which statements describe acceleration? Check all that apply.

Physics

1 answer:

Ad libitum [116K]2 years ago

3 0

Answer:

Negative acceleration occurs when an object slows down in the positive direction.

Negative acceleration occurs when an object speeds up in the negative direction.

Positive acceleration occurs when an object speeds up in the positive direction.

Positive acceleration occurs when an object slows down in the negative direction.

Explanation:

Basically same sign is speeding up and different sign is slowing down

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Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

2 years ago

The waves used by a microwave oven to cook food have a frequency of 2.45 gigahertz calculate the wavelength of this type of wave

fenix001 [56]

Electromagnetic wave bc I studied that early in the year

4 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not _____. This means the total m

drek231 [11]

Answer:

The law of conservation of momentum states that the total momentum of interacting objects does not change. This means the total momentum before a collision or explosion is equal to the total momentum after a collision or explosion.

6 0

2 years ago

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A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take

Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

$T =2\pi \sqrt{\frac{L}{g}}$

Here,

L = Length

g = Acceleration due to gravity

We can realize that $2 \pi$ is a constant so it is proportional to the square root of its length over its gravity,

$T \propto \sqrt{\frac{L}{g}}$

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

$g \rightarrow 0 \Rightarrow T \rightarrow \infty$

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

5 0

2 years ago

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$