Diego is trying to lift a piano to the second floor of his house. Diego uses a pulley system and gives a big lift to the piano.T
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This question is incomplete, the complete question is;
A woman is 160cm tall. What is the minimum vertical length of a mirror in which she can see her entire body while standing upright.
Hint: Consider the ray diagram below, of the rays that enable her to see her feet and the top of her head.
Use what you know about the law of reflection, together with a bit geometry.
The missing Image is uploaded along this answer below.
Answer:
the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm
Explanation:
Given the data in the question and illustrated in the image below,
From image 2;
The distance from the woman's eyes to the top of her head is represented as b and a represent the distance from her eyes to her feet.
Therefore, since her height is 160 cm
a + b = 160 ------ let this be equation 1
i.e AD + DG = 160 cm
Now, from the same image 2, we will notice that triangle ABC and tringle CBD are similar, so
∠ABC = ∠CBD
AC = CD
since AD = a and AC + CD = A
AC = CD = a/2
Also, triangle DEF and FEG are si,ilar
∠DEF = ∠FEG
so
DF = FG
since DG = b and DF + FG = b
DF = b/2
so the minimum vertical length of a mirror in which the woman can see her entire body while standing upright will be;
⇒ a/2 + b/2
⇒ a + b / 2
from equation 1, a + b = 160
so
⇒ a + b / 2 = 160 / 2 = 80
Therefore, the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm
Answer:
a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.
Explanation:
Statement is incomplete. The complete description is now described below:
<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>
<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>
<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>
Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (), measured in meters per second, is:
Where:
- Initial velocity in +x direction, measured in meters per second.
- Acceleration in +x direction, measured in meter per square second.
- Time, measured in seconds.
- Velocity in +y direction, measured in meters per second.
If we know that ,
,
and
, the final velocity of the satellite is:
a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:
The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:
The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.