The speed you read on a speedometer is ____.
1 answer:
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Answer:
It is real, inverted, and smaller than the object.
Explanation:
First of all, we can use the lens equation to find the location of the image:
where
q is the distance of the image from the lens
f = 15 cm is the focal length (positive for a converging lens)
p = 50 cm is the distance of the object from the lens
Solving the equation for q,
The distance of the image from the lens is positive, so we can already conclude that the image is real.
Now we can also write the magnification equation:
where are the size of the image and of the object, respectively.
Substituting p = 50 cm and q = 21.3 cm, we have
So from this relationship we observe that:
--> this means that the image is smaller than the object, and
--> this means that the image is inverted
so, the correct answer is
It is real, inverted, and smaller than the object.
Answer:
d) v = 100.2 mph, e) t = 1.25 s, f) -0.0340
Explanation:
d) This is an exercise that we can solve using projectile launch equations, let's start by calculating the time the ball will take to take home-plate
.x = 147 t
t = x / 147
t = 60.5 / 147
t = 0.41156 s
Let's use Pythagoras' theorem to find the speed
v = √ vₓ² + ²
vₓ = dx / dt
vₓ = 147
v_{y} = dy / dt
v_{y} = 4 -16 t
We look for speed for the time of arriving at home
= 4 - 16 0.41156
v_{y} = -2,585 ft / s
v_{y} = -2.585 ft/ s ( 1 mile /5280 foot) (3600s/1h)
Let's calculate the speed
v = √ (147² + 2,585²)
v = 147.02 ft / s
v = 147.0 ft/s (1 mile/5280 feet)(3600s/1h)
v = 100.2 mph
e) the time it takes for the ball to reach the floor and = 0 foot
y = 5 + 4 t - 16 t²
0 = 5 + 4t - 16t²
t² –t / 4 -5/4 = 0
t² -0.25 t -1.25 = 0
We solve the equation and second degree
t = [0.25 ±√(0.25² + 4 1.25)] / 2
t = [0.25 ± 2.25] / 2
t₁ = 1.25 s
t₂ = -1 s
The positive time is correct
t = 1.25 s
f) The angle of speed when the ball passes home
tan θ = / vₓ
θ = tan⁺¹ (v_{y} / vx)
The distance x is given in the exercise
x = 60.5 foot
vₓ = 147 foot / s
The speed y is t = 1.25 s
v_{y} = 5 + 4 1.25 - 16 1.25²
v_{y} = -15 foot / s
θ = tan⁻¹ (-15/147)
θ = -1,947º = -0,0340 rad