Answer:
Object should be placed at a distance, u = 7.8 cm
Given:
focal length of convex lens, F = 16.5 cm
magnification, m = 1.90
Solution:
Magnification of lens, m = -
where
u = object distance
v = image distance
Now,
1.90 = 
v = - 1.90u
To calculate the object distance, u by lens maker formula given by:
u = 7.8 cm
Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.
The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.
<h3>What is Geometry?</h3>
It deals with the size of geometry, region, and density of the different forms both 2D and 3D.
Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).
If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.
More about the geometry link is given below.
brainly.com/question/7558603
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Answer:

Explanation:
From the question we are told that:
Height of window 
Height of window off the ground 
Time to fall and drop
Generally the Newton's equation motion is mathematically given by

Where



Generally the Newton's equation motion is mathematically given by

Where





Therefore the ball’s initial speed

Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake