C. 400 m/s south is velocity (a vector quantity) as it gives both the magnitude as well as the direction of motion.
162.3 meters
Draw a diagram for the problem. You will
have a right triangle with a base of 75.0 m and an angle of 65.0
degrees. You want the opposite side of the triangle x. Using the law of
tangents, you have the following
x/75.0 = tan(65)
x = 75.0 * tan(65)
x = 75.0 * 2.144507
x = 160.8 meters
So now you know that the rocket went 160.8 meters above your eye. But
your eye isn't at ground level, it's 1.50 meters higher. So add 1.50
meters to the 160.8, giving you 162.3 meters.
Im pretty sure A-10 b-0 c-40
Answer:
q = 224 mm, h ’= - 98 mm, real imagen
Explanation:
For this exercise let's use the constructor equation
where f is the focal length, p and q are the distance to the object and the image respectively.
In a mirror the focal length is
f = R / 2
indicate us radius of curvature is equal to the diameter of the eye
R = 3,50 10² mm
f = 3.50 10² /2 = 1.75 10² mm
they also say that the distance to the object is p = 0.800 10³ mm
1 / q = 1 / f - 1 / p
1 / q = 1 / 175 - 1 /800
1 / q = 0.004464
q = 224 mm
to calculate the size let's use the magnification ratio
m =
h '=
h ’= - 224 350 / 800
h ’= - 98 mm
in concave mirrors the image is real.