What radiation do remote controls use?
1 answer:
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a) 50.8 mm
b) The whole image (1:1)
c) It seems reasonable
Explanation:
a)
To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:
where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem:
f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)
p = 3.00 m is the distance of the person from the lens
Therefore, we can find q:
b)
Here we need to find the height of the image first.
This can be done by using the magnification equation:
where:
y' is the height of the image
y = 1.75 m is the height of the real person
q = 50.8 mm = 0.0508 m is the distance of the image from the lens
p = 3.00 m is the distance of the person from the lens
Solving for y', we find:
(the negative sign means the image is inverted)
Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.
c)
This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.
Answer:
a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s
b) The time it takes the car to reach the edge is 4.79 s
c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal
d) The total time interval the car is in motion is 6.34 s
e) The car lands 24 m from the base of the cliff.
Explanation:
Please, see the figure for a description of the situation.
a) The equation for the position of an accelerated object moving in a straight line is as follows:
x =x0 + v0 * t + 1/2 a * t²
where:
x = position of the car at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:
x = 1/2 a * t²
With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.
46.5 m = 1/2 * 4.05 m/s² * t²
2* 46.5 m / 4.05 m/s² = t²
<u>t = 4.79 s </u>
The equation for velocity is as follows:
v = v0 + a* t
Where:
v = velocity
v0 = initial velocity
a = acceleration
t = time
For the car, the velocity will be
v = a * t
at the edge, the velocity will be:
v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>
b) The time interval was calculated above, using the equation of the position:
x = 1/2 a * t²
46.5 m = 1/2 * 4.05 m/s² * t²
2* 46.5 m / 4.05 m/s² = t²
t = 4.79 s
c) When the car falls, the position and velocity of the car are given by the following vectors:
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
v =(v0x, v0y + g * t)
Where:
r = position vector
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity
v = velocity vector
First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.
Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:
cos -37.0º = v0x / v0
(the angle is negative because it was measured clockwise and is below the horizontal)
(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)
v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s
sin 37.0º = v0y/v0
v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s
Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):
y = y0 + v0y * t + 1/2 * g * t²
Notice in the figure that the y-component of the vector "r final" is -30 m, then:
-30 m = y0 + v0y * t + 1/2 * g * t²
According to our reference system, y0 = 0:
-30 m = v0y * t + 1/2 g * t²
-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²
0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²
Solving this quadratic equation:
<u>t = 1.55 s</u> ( the other value was discarded because it was negative).
Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:
vy = v0y + g * t
vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s
The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).
The velocity vector when the car lands is:
v = (15.5 m/s, -26.9 m/s)
We have to express it in magnitude and direction, so let´s find the magnitude:
To find the direction, let´s use trigonometry again:
sin α = vy / v
sin α = 26.9 m/s / 31.0 m/s
α = 60.2º
(notice that the angle is measured below the horizontal, then it has to be negative).
Then, the vector velocity expressed in terms of its magnitude and direction is:
vy = v * sin -60.2º
vx = v * cos -60.2º
v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)
<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>
d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.
total time = falling time + rolling time
total time = 1,55 s + 4.79 s = <u>6.34 s</u>
e) Using the equation for the position vector, we have to find "r final 1" (see figure):
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
Notice that the y-component is 0 ( figure)
we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.
r final 1 = ( v0x * t, 0)
r final 1 = (15.5 m/s * 1.55 s, 0)
r final 1 = (24.0 m, 0)
<u>The car lands 24 m from the base of the cliff.</u>
PHEW!, it was a very complete problem :)
