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kenny6666 [7]
2 years ago
13

NEED answer pls ,dont understand hope someone can help it is multiple choice except 12,13,14,15

Mathematics
1 answer:
myrzilka [38]2 years ago
5 0

Answer:

C.) Corresponding; B.) m<6 = m<2; and for the last one you probably have to divide so perhaps it might be 30 .

Step-by-step explanation:

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A mass of 1 g is set in motion from its equilibrium position with an initial velocity of 6in/sec, with no damping and a spring c
yan [13]

a) y(t)=0.0016 sin(94.9t) [m]

b) 0.033 s

c) -0.152 m/s

Step-by-step explanation:

a)

The force acting on the mass-spring system is (restoring force)

F=-ky

where

k = 9 is the spring constant

y is the displacement

Also, from Newton's second law of motion, we know that

F=my''

where

m = 1 g = 0.001 kg is the mass

y'' is the acceleration

Combining the two equations,

my''=-ky

This is a second order differential equation; the solution for y(t) is

y(t)=A sin(\omega t-\phi)

where

A is the amplitude of motion

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{9}{0.001}}=94.9 rad/s is the angular frequency

The spring starts its motion from its equilibrium position, this means that y=0 when t=0; therefore, the phase shift must be

\phi=0

So the displacement is

y(t)=A sin(\omega t)

The velocity of the spring is equal to the derivative of the displacement:

v(t)=y'(t)=\omega A cos(\omega t)

We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,

v_0=6(0.0254)=0.152 m/s

And since at t = 0, cos(\omega t)=1

Then we have:

v_0=\omega A

From which we find the amplitude:

A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m

So the solution for the displacement is

y(t)=0.0016 sin(94.9t) [m]

b)

Here we want to find the time t at which the mass returns to equilibrium, so the time t at which

y=0

This means that

sin(\omega t)=0

We know already that the first time at which this occurs is

t = 0

Which is the beginning of the motion.

The next occurence of y = 0 is instead when

\omega t = \pi

which means:

t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s

c)

As said in part a), the velocity of the mass-spring system at time t is given by the derivative of the displacement, so

v(t)=\omega A cos(\omega t)

where we have

\omega=94.9 rad/s is the angular frequency

A=0.0016 m is the amplitude of motion

t is the time

Here we want to find the velocity of the mass when the time is that calculated in part b):

t = 0.033 s

Substituting into the equation, we find:

v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s

4 0
3 years ago
Find x and y in the kite
marin [14]
The 3rd option, x=75y=35.
7 0
2 years ago
Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for
USPshnik [31]

x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}

\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}

6 0
3 years ago
A trimming operation at a local manufacturer produces rods whose length conforms to unifrom distribution with a minimum of 18cm
MariettaO [177]

Answer:

The probability that randomly selected road will be at least 25.8 cm long will be 48%.

Step-by-step explanation:

Given: uniform distribution with min and max values of 18 and 33 respectively.

To find : probability density for upper cumulative frequency i.e 25.8 at least means x\geq25.8  upto 33 cm i.e the maximum limit of function.

Solution:

we have by definition , of uniform distribution

we get , probability density function defines as :

<em>F(x,a,b)= </em>\frac{1}{b-a}<em>   </em>a\leq x\leq b

            =1/(33-18)=1/15=0.0667.

this is probability density function.

here the x=25.8 , a=18 and b=33

for lower cumulative frequency it defines as ;

P(x,a,b)=\frac{x-a}{b-a} =25.8-18/33-18=0.52

for upper cumulative frequency it defines as ;

Q(x,a,b)=b-x/b-a=33-25.8/33-18=0.48

here at least 25.8 cm probability means it should be greater than a value(18cm) hence it is provided by the upper cumulative frequency

i.e. Q(x,a,b)=0.48

The probability that randomly selected road will be at least 25.8 cm long will be 48%.

7 0
2 years ago
Given h(x) = 5x -1, find h(1).
Norma-Jean [14]
H(1) = 4


h(1)= 5 x 1 -1
=5-1
=4
hope that helps <3
5 0
3 years ago
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