The particle's acceleration is 5.1 m/s²
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What is Acceleration ?</h3>
Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²
Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s
The given parameters are;
Acceleration a = ΔV ÷ ΔT
a = (2.35 + 8.72) / (5.59 - 3.42)
a = 11.07 / 2.17
a = 5.1 m/s²
Therefore, the particle's acceleration is 5.1 m/s²
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Speed =distance/time
3.25=3.00/time
3.25xt=3.00
t=3/3.25
s=0.9s
Answer:
The depth of focus achievable with those lenses is small.
Explanation:
A larger aperture makes it much harder to focus on more than one object. When using a telephoto lens (the ones the question is referring to), the depth of focus is very small. For example, using a telephoto lens to take a photo of a runner might get the runner in focus, but certainly not the track, or the audience behind them. If you look at photos, especially older photos, of Olympians in almost any sport you can see this.
Hope this helps!
I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
