Answer: condensation.
Vaporization is the pass from liquid state to gaseous state.
Then the reverse is the transformation from gaseous state to liquid state.
That is called condensation.
When the water vaporizes the liquid transforms into vapor which goes to the atmosphere. When the water vapor of the atmosphere condensates liquid water is formed. You can see condensation when you have a glass with cold water and drops of water form in the exterior of the glass: those drops are liquid water that formed when the vapor of the air that surrounds the glass cools due to the lower temperature of the surface of the glass.
We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.
Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s
<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
Lunar phase is the same wherever on Earth you observe
<span>Last (third) quarter rises at midnight, sets at noon. </span>
<span>First quarter rises at noon, sets at midnight</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)
