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3 years ago

6

Each day, Ted can wax 8 cars or wash 10 cars, and Tom can wax 3 cars or wash 5 cars. What is each man's opportunity cost of wash

ing a car?

1 answer:

valentinak56 [21]3 years ago

3 0

Answer:

1/2

Explanation:

This will lead us to simultaneous equation since there are two person involved doing the same job each day.

For Ted;

Since Ted can wax 8 cars or wash 10 cars in a day. Mathematically, we have

8x + 10w = 1... (1)

Where x is for waxing cars, w is for washing cars

Similarly for Tom,

Tom can wax 3 cars or wash 5 cars in a day as well, this gives us;

3x + 5w = 1... (2)

Equating 1 and 2, we have;

8x + 10w = 1... (1)

3x + 5w = 1... (2)

Using elimination method, we will multiply eqn 1 by 3 and eqn 2 by 8 to have;

24x + 30w = 3

24x + 40w = 8

Subtracting eqn 4 from 5 we have;

30w - 40w = 3 - 8

-10w = -5

w = 5/10

w = 1/2

Therefore each man's opportunity cost of washing a car is 1/2

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TEA [102]

Answers:

a) 122.5 m

b) 50 m

Explanation:

Here we are dealing with projectile motion, where we can use the following two equations that model the parabolic motion of the stone after being thrown from the window:

$h=h_{o}+V_{o}sin(\theta)t-\frac{1}{2}gt^{2}$ (1)

$x=V_{o}cos(\theta)t$ (2)

Where:

$h=0 m$ is the height of the stone when it hits the ground

$h_{o}$ is the initial height of the stone

$V_{o}=10 m/s$ is the initial velocity of the stone

$\theta=0\°$ since we are told the stone was thrown horizontally

$g=9.8 m/s^{2}$

$t=5 s$ is the time the stone is in air

$x$ is the horizontal distance the stone travelled

Knowing this, we can answer the questions:

a)Initial height of the stone:

In this case we will isolate $h_{o}$ from (1):

$h_{o}=\frac{1}{2}gt^{2}$ (3)

$h_{o}=\frac{1}{2}(9.8 m/s^{2})(5 s)^{2}$ (4)

$h_{o}=122.5 m$ (5) This is the initial height

b) Horizontal distance:

In this case we will use equation (2):

$x=(10 m/s)cos(0\°)(5 s)$ (6)

$x=50 m$ (7) This is the horizontal distance

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A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

QveST [7]

Answer:

$\Delta \theta = 47.57^{\circ} C$

Explanation:

given,

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$\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}$

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we know,

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now,

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$\Delta \theta = 47.57^{\circ} C$

the temperature change per compression stroke is equal to 47.57°C.

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