Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
Find the velocity of the object after one second.
v = vo + at
v = (0 m/s) + (9.8 m/s^2)(1 s)
v = 9.8 m/s
Now, using that, you can find the displacement in that one second between 1 and 2.
d = vot + (1/2)at^2
d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2
d = 14.7 m
Given what we know, we can confirm that due to its relatively low image quality, any memory card with at least one MB will be sufficient to hold the 120 images taken.
<h3>Why is one MB enough?</h3>
The photos are taken at 8 megapixels, with each one occupying 14 bits of information. A megabit consists of one million bits of potential storage. Therefore, the 120 pictures will occupy roughly 1% of the available storage of a 1 MB memory card.
Therefore, we can confirm that due to its relatively low image quality, any memory card with at least one MB will be sufficient to hold the 120 images taken.
To learn more about megabits visit:
brainly.com/question/22735284?referrer=searchResults
Answer:
ΔP = 20000 N s
Explanation:
To solve this problem we use the relation between momentum and moment
I = Δp
let's calculate the momentum
I = ∫F dt
if we use the average force
I = F t
I = 10000 2
I = 20000 N s
therefore with the first equation
ΔP = I = 20000 N s
