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3 years ago

13

He cartesian coordinates of a point are given. (a) (−9, 9) (i) find polar coordinates (r, θ) of the point, where r > 0 and 0

≤ θ < 2π.

1 answer:

olga_2 [115]3 years ago

5 0

Radius =
9^2 + 9^2 = d^2
d = sqrt(162), about 12.73

theta =
atan(9/-9)

is 45°, but in 2nd quad, so add 90

(12.727, 135°)

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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:

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

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3 years ago

Order the decimals from least to greatest 3.4897, 3.3487, 1.5467, 3.348

Anna71 [15]

Answer:

1.5467, 3.348, 3.3487, 3.4897

Step-by-step explanation:

Remember to compare each one number at a time!

For the first number, only one number is 1 while the others are 3 making it the lowest.

Then when comparing 3.348 and 3.3487, it can be implied that there is a 0 at the end of 3.348 because it makes the numbers have the same # of decimal places without changing the value of the number (3.3480). That number is less than 7.

Finally, 3.4897 has a 4 in the tenths place instead of a 3 making it the highest number.

Hope this helps!

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3 years ago

Find the discount on a note for $23,000 made by Jacques on February 10 to be repaid on August 10 of the same year with a discoun

Olenka [21]

Total Value =$23,000

Time = 10 february to 10 august = 6 months

And

Rate of discount = 9% =0.09

Now we have to find the discount on the note

So we multiply the amount , time and rate

but time should be in years

So 6 months = 6/12 years = 0.5 years

So

Discount = 23,000 * 0.5 * 0.09 = 1035

Hence total Discount = $1035

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3 years ago

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A fish is 3 feet below the surface of a lake. If its position can be recorded as −3 feet, what would the position of 0 represent

trasher [3.6K]

Answer:

On the surface of the lake

Step-by-step explanation:

A position of 0 represent that the object is on the surface of the lake. Negative 3 represents 3 feet below the surface of the lake.

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