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3 years ago

7. A car travels at 25 m/s to the North. It has an acceleration of 2 m/s to the south

Physics

1 answer:

tamaranim1 [39]3 years ago

7 0

Answer:

-0.5 m/s2 as it comes to rest.

Explanation:

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aivan3 [116]

Answer:

same

Explanation:

8 0

2 years ago

Read 2 more answers

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

3 years ago

Question 6 (10 points)

horsena [70]

Answer:

a= g = - 9.81 m/s2.

The following equations will be helpful:

a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad

When you substitute the specific acceleration due to gravity (g), the equations are as follows:

g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd

If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:

g = vf /t d = 1/2 gt2 vf2 = 2gd

The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.

hope this info helps you out!

7 0

3 years ago

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp

laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

$a = \frac{v^2}{r}$

At constant speed, we will have;

$v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1$

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0

3 years ago

Define what is boyleâs law?

lawyer [7]

Answer:

Volume of given mass of gas is inversely proportional to pressure of gas

Explanation:

Boyle's law: It states that the volume of given mass of gas is inversely proportional to the pressure of gas at constant temperature.

Mathematical representation:

Suppose, a gas of mass m

T=Constant temperature

V=Volume of gas

P=Pressure of gas

Then, $V\propto\frac{1}{P}$

8 0

3 years ago