Answer:
3. you need to ask your available lab instructors what to do.
4. You immediately have to drop down your cloth and roll it to extinguish the fire or move to the emergency shower if available
5. You have too keep calm and report to the lab instructor but do no shout.
6. Move immediately to the eye rinse basin if available and wash your eyes gently and thoroughly
Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
Acceleration = (change in speed) / (time for the change)
change in speed = (ending speed) - (starting speed)
change in speed = (10 m/s) - (2 m/s) = 8 m/s
Acceleration = (8 m/s) / (4 sec)
Acceleration = (8/4) (m/s²)
<em>Acceleration = 2 m/s²</em>
For the given question above, I think there is an associated choice of answer for it. However, the answer for this is London Dispersion Forces. <span>Dipole-dipole forces and hydrogen bonding are much stronger, leading to higher melting and boiling points.</span>
Answer:
W = 2352 J
Explanation:
Given that:
- mass of the bucket, M = 10 kg
- velocity of pulling the bucket, v = 3

- height of the platform, h = 30 m
- rate of loss of water-mass, m =

Here, according to the given situation the bucket moves at the rate,

The mass varies with the time as,

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance
∆x = 3∆t meters
So, during this interval change in work done,
∆W = m.g∆x
<u>For work calculation:</u>
![W=\int_{0}^{10} [(10-0.4t).g\times 3] dt](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B10%7D%20%5B%2810-0.4t%29.g%5Ctimes%203%5D%20dt)
![W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}](https://tex.z-dn.net/?f=W%3D%203%5Ctimes%209.8%5Ctimes%20%5B10t-%5Cfrac%7B0.4t%5E%7B2%7D%7D%7B2%7D%5D%5E%7B10%7D_%7B0%7D)
