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Marta_Voda [28]

3 years ago

5

Yvonne has gained a reputation for her cheesecakes. She takes orders for 10 cheesecakes and spends 8.5 hours on one Saturday bak

ing them all. She earns $120 from selling these cheesecakes. a. What is the average length of time to make one cheesecake? Give your answer in minutes. b. What does Yvonne charge for each cheesecake? c. How much money is Yvonnne earning per hour for her work?

1 answer:

VLD [36.1K]3 years ago

3 0

A) 10÷8.5
b) 120÷10
c) answer A÷Answer B.

sure on first two, not so sure on C.

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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

makkiz [27]

Answer:

Explanation:

$\lambda$ = Observed wavelength = 550 nm

$\lambda'$ = Actual wavelength = 635 nm

c = Speed of light = $3\times 10^8\ \text{m/s}$

v = Velocity of the physicist

Doppler shift is given by

$f=\sqrt{\dfrac{c+v}{c-v}}f'\\\Rightarrow \dfrac{c}{\lambda}=\sqrt{\dfrac{c+v}{c-v}}\dfrac{c}{\lambda'}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{c+v}{c-v}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}-1=\dfrac{v}{c}(1+\dfrac{\lambda'^2}{\lambda^2})\\\Rightarrow v=\dfrac{c(\dfrac{\lambda'^2}{\lambda^2}-1)}{1+\dfrac{\lambda'^2}{\lambda^2}}$

$\Rightarrow v=\dfrac{3\times 10^8\times (\dfrac{635^2}{550^2}-1)}{1+\dfrac{635^2}{550^2}}\\\Rightarrow v=42817669.77\ \text{m/s}$

The physicist was traveling at a velocity of $42817669.77\ \text{m/s}$ .

4 0

3 years ago

A flashlight bulb is connected to two dry cells with an equivalent voltage of 4.50 V. If the bulb draws a current of 12.0 mA (10

trasher [3.6K]

Answer:

b. 375 Ohms

Explanation:

<u>Given the following data;</u>

Voltage = 4.5 V

Current = 12 mA

To find the resistance;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$

Where;

V represents voltage measured in voltage.
I represents current measured in amperes.
R represents resistance measured in ohms.

<u>Conversion:</u>

1000 mA = 1 A

12 mA = 12/1000 = 0.012 A

$R = \frac {V}{I}$

Substituting into the formula, we have;

$R = \frac {4.5}{0.012}$

Resistance, R = 375 ohms

6 0

3 years ago

An astronomer sees two stars in the sky. Both stars are equally far from Earth, but the first star is dimmer than the second sta

kvv77 [185]

Answer: It is smaller than the second star

Explanation: The brightness of a star depends on two factors: The distance from the Earth and size of the star. More the distance from the Earth, the star appears dimmer. And bigger the star, brighter it is.

$b =\frac{L}{4\pi d^2}$

Where L is the luminosity,d is the distance

Luminosity depends on the surface area of star $(4\pi R^2)$

Here, it is given that the two stars are at the same distance from the Earth. The first star is dimmer. We can conclude from this that first star star must be smaller than the second star.

7 0

3 years ago

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At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in sim

Rashid [163]

T = 12.4 h, x(t) = .250 d (displacement)
The total amplitude is half of the distance d from the highest to lowest level.
Using the equation x(t) = x(m) cos(wt + φ), we get:

.250d = .5d cos(wt + φ)
To find w, use the equation w = 2π / T

.250d = .5d cos((2π / 12.4)t + φ)

.250d = .5d cos((2π / 12.4)t + 0)

Solve the equation for t.
.250d / .5d = cos ((2π / 12.4) t)
arccos(.5) = 2π / 12.4 * t
arccos(.5) * 12.4 / 2 / π = t
t = 2.066666 hours

8 0

3 years ago