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3 years ago

Find the momentum of a 25 kg object traveling at a speed of 4 m/s

Physics

1 answer:

White raven [17]3 years ago

4 0

Answer:

100 kg*m/s

Explanation:

P = m*v

= 25*4 = 100

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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

A 3.53-g sample of aluminum completely reacts with oxygen to form 6.67 g of aluminum oxide. find percent composition

katrin2010 [14]

The mass percent composition of aluminum is 52.9% in aluminum oxide.

Mass of the aluminum = 3.53 g

Mass of the aluminum oxide = 6.67 g.

The mass percent of a substance is the mass of the substance divided by the mass of the compound into 100.

Aluminum reacts with oxygen to form aluminum oxide.

The overall balanced equation for the reaction is,

$Aluminum + Oxygen→Aluminum \: oxygen$

$4Al +3O _{2}→2Al _{2}O _{3}$

The mass percent composition of aluminum in the aluminum oxide is,

$Mass \: percent = \frac{ Mass \: of \: the \: substance}{ Mass \: of \: the \: compound} \times 100$

$Mass \: percent = \frac{Mass \: of \: the \: aluminum}{ Mass \: of \: the \: aluminum \: oxide } \times 100$

$Mass \: percent = \frac{3.53}{6.67 } \times 100$

= 52.9 %

Therefore, the mass percent composition of aluminum is 52.9% in aluminum oxide.

To know more about aluminum oxide, refer to the below link:

brainly.com/question/25869623

#SPJ4

7 0

1 year ago

Select the correct answer.

Vesna [10]

Answer:

Explanation:

Radiant= list onto solar panels, Electric= solar into power, Radiant= Electric into light

8 0

2 years ago

Read 2 more answers

Force F → = (−8.0 N)iˆ + (6.0 N)jˆ acts on a particle with position vector r → = (3.0 m)iˆ + (4.0 m)jˆ. What are (a) the torque

disa [49]

Answer with Explanation:

We are given that

$F=-8\hat{i}+6\hat{j}$

$r=3\hat{i}+4\hat{j}$

a.We have to find the torque on the particle about the origin.

We know that

Torque= $\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}$

By using the formula

$\tau=50\hat{k}$

b. $\mid \tau\mid =\mid F\mid \mid r\mid sin\theta$

$\mid F\mid=\sqrt{(-8)^2+(6)^2}=10$

$\mid r\mid=\sqrt{3^2+4^2}=5$

$\mid \tau\mid=\sqrt{(-50)^2}=50$

Substitute the values then we get

$50=10\times 5 sin\theta$

$sin\theta=\frac{50}{50}=1$

$sin\theta=sin90^{\circ}$

Because $sin90^{\circ}=1$

$\theta=90^{\circ}$

3 0

3 years ago

The relationship between an object's acceleration and its mass

Shalnov [3]

Answer:

Newton's second law

Explanation:

The relationship between mass and acceleration is described in Newton's Second Law of Motion. His Second Law states that the more mass an object has, more force is necessary for it to accelerate.

5 0

3 years ago