Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s
Answer:
1 / f = 1 / p + 1 / q
Explanation:
For this exercise we will use two methods, an analytical method which is to use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, positive for converging lenses and negative for diverging lenses, p and q are the distance to the object and the image respectively
m = h’/ h = - q / p
where m is the magnification, h ’and h are the height of the image and the object.
We will also use a graphical method, where three rays will be traded
1) A ray that passes through the center of the lens and should not
2) a ray that passes through the focal length and comes out parallel to the lens
3) A ray that is horizontal and comes out through the focal from the other side
for this second method see the attachment
Answer:
The mass of the rod is 16 kg.
Explanation:
Given that,
The length of a rod, L = 3 m
The moment of inertia of the rod, I = 12 kg-m²
We need to find the mass of the rod. The moment of inertia of the rod of length L is given by :
Where
M is mass of the rod
So, the mass of the rod is 16 kg.
<span>The answer here is positive space. Within a composition, any given object occupies positive space, whereas white space around this, or simply areas of blank space that are not being somehow used, is defined as negative space. Hopefully this clears up your question!</span>
(a) We must first look at the formulas of the velocities of each toy car. v1 =
-4.2 + 2.60t. v2 = 5.20. When the two cars have equal speed, then
v1 = v2
-4.2 + 2.60t = 5.20
2.60t = 9.40
t = 3.62 s
(b) Their speed would then be 5.20 m/s. The toy car does not change speed since it doest not have any acceleration.
(c) The two cars will pass each other when their positions are equal.
x1 = 13.5 - 4.2t + 0.5*2.60t^2
x2 = 8.5 + 5.20t
x1 = x2
13.5 - 4.2t + 1.30t^2 = 8.5 + 5.20t
1.30t^2 - 9.40t + 5.0 = 0
t = 6.65s or t = 0.58 s
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