A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i
mage distance and the image size. Also indicate the orientation of the image compared to the object as well as if it will be a real or virtual image.
1 answer:
Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.
Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:
1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.
replacing the values we obtain:
1/8.3+1/q=1/15.2
so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3
then q=-18.28 cm
The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2
We also add a picture to see the imagen formation for this case.
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Answer:
The ball will drop 0.881 m by the time it reaches the catcher.
Explanation:
The position of the ball at time "t" is described by the position vector "r":
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
When the ball reaches the catcher, the position vector will be "r final" (see attached figure).
The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.
Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):
x = x0 + v0x · t
17.0 m = 0 m + 40.1 m/s · t
t = 17.0 m/ 40. 1 m/s = 0.424 s
With this time, we can calculate the y-component of the vector "r final", the drop of the ball:
y = y0 + v0y · t + 1/2 · g · t²
Initially, there is no vertical velocity, then, v0y = 0.
y = 1/2 · g · t²
y = -1/2 · 9.8 m/s² · (0.424 s)²
y = -0.881 m
The ball will drop 0.881 m by the time it reaches the catcher.
