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3 years ago

Charlie Brown kicks a football at 24.5 m/s at 35.0. What is the maximum height of the ball?

Physics

1 answer:

lozanna [386]3 years ago

5 0

Answer:

d = 10.076 m

Explanation:

We need to obtain the velocity of the ball in the y direction

Vy = 24.5m/s * sin(35) = 14.053 m/s

To obtain the distance, we use the formula

vf^2 = v0^2 -2*g*d

but vf = 0

d = -vo^2/2g

d = (14.053)^2/2*(9.8) = 10.076 m

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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

3 years ago

Two charges with a certain distance apart have an electrostatic force of 400 N acting

Doss [256]

As the charges’ distance increase, there is a weaker force of attraction between them hence the electrostatic force decreases as distance increases. It increase by 4 (times 4) so the force will decrease by 4 making the answer

=A (400 divided by 4 = 100)

3 0

2 years ago

An automobile moves at a constant speed over the crest of a hill traveling at a speed of 88.5 km/h. At the top of the hill, a pa

Sergio039 [100]

Answer:

$r=61.65m$

Explanation:

Since the package remains in contact with the car's seat, the package's speed is equal to the car's speed. At the top on the mountain the package's centripetal force must be equal to its weight:

$mg=F_c$