Answer:-2.86*10⁻⁴
Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B
Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴
ΔP = P₂-P₁ ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷
<span>All the statements about the atmosphere are correct except statement 2. The atmosphere has effects on landforms and does affect the biosphere. For instance, landforms are constantly been broken down and renew by the process of weathering, erosion, etc. The atmosphere plays principal roles in these processes. </span>
<span>If the temperature increases in a sample of gas at constant volume, then its pressure increases. The increase in temperature makes the molecule hit the walls of the container faster. The correct option among all the options that are given in the question is the third option or option "c". I hope the answer helps you.</span>
Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880
Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000
Lost Energy = 5880 - 3000 = 2880 J
In short, Your Answer would be 2880 Joules
Hope this helps!
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)
