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LenKa [72]
3 years ago
13

A source of laser light sends rays AB and AC toward two opposite walls of a hall. The light rays strike the walls at points B an

d C, as shown below:
A source of laser light is at point A on the ground between two parallel walls. The walls are perpendicular to the ground. AB is a ray of light which strikes the wall on the left at point B. The length of AB is 40m. AC is a ray of light which strikes the wall on the right at point C which is 80 m above the ground. The ray AB makes an angle of 60 degrees with the ground. The ray AC makes an angle of 45 degrees with the ground.

What is the distance between the walls?
Mathematics
1 answer:
snow_tiger [21]3 years ago
3 0

The distance between the walls is: 76.56854249

Cos(60)=A/40

Cos(45)=A/80

20+56.56854249

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A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co
butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

\displaystyle \vec{v}=\frac{\vec{x}}{t}

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\displaystyle \vec{x}=\vec{v}.t

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=\ km/h

The displacement of the plane in 2 hours is

\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)

\displaystyle \vec{x_1}=km

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=km/h

The displacement in 1 hour is

\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)

\displaystyle \vec{x_2}=km

The total displacement is the vector sum of both

\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+

\displaystyle \vec{x_t}=km

\displaystyle \vec{x_t}=

The distance from the airport is the module of the displacement:

\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}

\displaystyle |\vec{x_t}|=894.43\ km

8 0
4 years ago
Line segments JK and JL in the xy-coordinate plan both have a common endpoint J(-4,11) and midpoints at M1 (2,16) and M2 (-3,5),
OleMash [197]
The first part of the question is not needed 
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6 0
4 years ago
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weqwewe [10]

Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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5 0
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