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MA_775_DIABLO [31]

3 years ago

15

A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of h

er original course, and flies 2 hours 15 minutes in the new direction. If she maintains a constant speed of 345 mi/h, how far is she from her starting position? Give your answer to the nearest mile.

1 answer:

mrs_skeptik [129]3 years ago

4 0

Answer:

1199 miles

Explanation:

1 hour 30 minutes = 1 + 30/60 = 1.5 hours

2 hours 15 minutes = 2 + 15/60 = 2.25 hours

The distance she flew in the 1st segment is:

1.5*345 = 517.5 miles

The distance she flew in the 2nd segment is:

2.25 * 345 = 776.25 miles

Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown

776.25 * cos(45) = 549 miles in-line with the 1st segment and

776.25* sin(45) = 549 miles perpendicular to the 1st segment:

So the distance from the end to her starting position is

$\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles$

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A tabletop gamer has designed a game that requires three dice to be thrown onto a tray with a measurement grid. To add an extra

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Answer:

The resulting coordinate is $(x,y) = (-0.170 \ m , -0.038 \ m )$

Explanation:

From the question we are told that

The mass of the first dice is $m_1 = 11.10 \ g$

The mass of the second dice is $m_2 = 15.10 \ g$

The mass of the third dice is $m_3 = 18.90 \ g$

The coordinate of the first dice is $(x_1, y_1) = (0.3150\ m , -0.4990 \ m )$

The coordinate of the second dice is $(x_2,y_2) = (-0.4050 \ m , 0.4850 \ m )$

The coordinate of the third dice is $(x_3 , y_3) = (-0.1150 \ m , -0.1850\ m )$

Generally the resulting coordinate of the center of mass of the dice in the x-axis is mathematically evaluated as

$x\ cm = \frac{m_1 * x_1 + m_2 * x_2 + m_3 * x_3}{m_1 + m_2 +m_3 }$

i.e the summation of the moments about their x-axis divided by the magnitude of their masses

substituting values

$x\ cm = \frac{11.0 * 0.3150 + 15.10 * (-0.4050) + 18.90 * (-0.1150)}{11.100 + 15.10 +18.90 }$

$x\ cm = -0.170 \ m$

Generally the resulting coordinate of the center of mass of the dice in the y-axis is mathematically evaluated as

$y\ cm = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3}{m_1 + m_2 +m_3 }$

$y\ cm = \frac{ 11.10 * (-0.4990) + (15.10) * (0.4850) + (18.90) * (-0.1850)}{ 11.10 + 15.10 +18.90 }$

$y\ cm = -0.038 \ m$

Thus the resulting coordinate is $(x,y) = (-0.170 \ m , -0.038 \ m )$

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3 years ago

The density of air at room temperature is about 1.2 g/L. This is the same as

Verdich [7]

The correct answer is
<span>C) 1200 g/m3.

Let's see why. The relationship between liters and cube decimeters is
</span> $1 L = 1 dm^3$
Therefore,
$1 g/L= 1 g/dm^3$
However, we also know that
$1 dm^3 = 10^{-3} m^3$
Therefore,
$1 L = 10^{-3} m^3$
and
$1 \frac{g}{L}= 1 \frac{g}{10^{-3} m^3} =1 \cdot 10^3 \frac{g}{m^3} =1000 \frac{g}{m^3}$

Therefore, the density of the problem $1.2 g/L$ becomes
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3 years ago

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Answer:

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$F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)$

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

$N=m*g*cos(\alpha)$

and the force of static friction (f) is given as the static coefficient of friction ( $\mu$ ) times the normal N:

$f=\mu *m*g*cos(\alpha)$

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

$f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)$

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

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3 years ago