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2 years ago

6

sandra lives in a country located in the northern hemisphere she makes a sundial by erecting a pole vertically in her garden the

given figure shows the aerial view of the pole and the shadow it casts at 6:00 am on a summer day which of these figures shows the shadow of the same pole at 6:00 pm on the same day

1 answer:

Arte-miy333 [17]2 years ago

5 0

Answer:

C

Explanation:

The sun has rotated causing the shadow to reflect.

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3 years ago

Write a linear equation that expresses the relationship between the temperature in degrees Celsius (C) and degrees Fahrenheit (F

UkoKoshka [18]

Answer:

$\displaystyle F=\frac{9}{5}C+32$

$\displaystyle C=\frac{5}{9}(F-32)$

Explanation:

<u>Temperature Units Conversion </u>

The conversion formula between Celsius and Fahrenheit temperature scales is well-known. But we'll use the provided data to derive the formula. Let's model the relationship between Fahrenheit (F) and Celsius (C) as a linear function like

$F=mC+b$

Where m and b must be computed according to the pair of conditions given. The values for each temperature scale are (C,F)=(0,32) and (100,212). Replacing the first value

$32=m\times 0+b$

It means that

$b=32$

By using the second point

$212=m\times 100+32$

Solving for m

$\displaystyle m=\frac{212-32}{100}=\frac{180}{100}$

Simplifying

$\displaystyle m=\frac{9}{5}$

So, the conversion formula is

$\displaystyle F=\frac{9}{5}C+32$

Which is the widely known formula for temperature conversion

Solving for C, we get the inverse relation

$\boxed{\displaystyle C=\frac{5}{9}(F-32)}$

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3 years ago

A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass

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To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

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

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

$KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2$

Where I is the moment of inertia previously defined.

$KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2$

In the case of the final kinetic energy, we have to,

$KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$ , then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)$

$2.2388= (\frac{7}{5}) (v)^2$

$v=1.26m/s$

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3 years ago