<u>First law of thermodynamics:</u>
- It states that <em>"Energy neither be created nor it can be destroyed". </em>simply it converts one form of energy into another form.
- It is also known as<em> "law of conservation of energy"</em>
<u>Limitations of First law</u>
- It doesn't provide a clear idea about the direction of transfer of heat.
- It doesn't provide the information that how much heat energy converted inti work.
- Its not given any practical applications.
<u>II law of thermodynamics:</u>
It states that <em>"the total entropy of the system can never decrease over time"</em>
It is strongly proved by two laws, they are
<em>1. Kelvin-plank statement:</em>
He stated that "any engine does not give 100% efficiency". It violates the Perpetual motion of machine II kind<em>(PMM-II).</em>
<em>2. Classius statement: </em>
<em> </em><em> It states that "Heat always flows from high temperature body to low temperature body, without aid of external energy". </em>
<em> Also it stated that " Heat can also be transferred from low temperature body to high temperature body, by the aid of an external energy".</em>
<em>Applications of II law: </em>
<em>Refrigeration &Air conditioning, Heat transfer, I.C. engines, etc.</em>
<span>A. </span>Let’s
say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the
components are v0x and v0y.
<span>v0y = 0 since the customer slides it horizontally so applied
force is in the x component only.
<span>The equations for horizontal and vertical projectile motion
are:
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>
Setting the origin to be the end corner of the
counter so that x0=0 and y0=0, hence:
x = v0x t
y = - 1/2 g t^2
Given value are: x=1.50m and y=-1.15m (y is
negative since mug is going down)
<span>1.50m = v0x t
----> v0x= 1.50/t</span>
<span>-1.15m = -(1/2) (9.81) t^2 -----> t =0.4842 s</span>
Calculating for v0x:
v0x = 3.10 m/s
<span>B. </span>v0x
is constant since there are no other horizontal forces so, v0x=vx=3.10m/s
vy can be calculated from the formula:
<span>vy = v0y + at where a=-g
(negative since going down)</span>
vy = -gt = -9.81 (0.4842)
vy = -4.75 m/s
Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy
tan(90-Ø )= 3.1/4.75
Ø =
56.87˚<span> below the horizontal</span>
Answer:
An object can have a displacement in the absence of any external force acting on it
Explanation:
When a object moves with a constant velocity (v), then it gets displaced in the direction of motion but the net external force experienced by the object is zero.
F external =ma
If object moves with constant velocity, acceleration is zero.
Since, a=0 ⟹F external =0
Using s=ut+ 1/2 at ^2
⟹ Displacement s=ut (∵a=0)
Hence, an object can have a displacement in the absence of any external force acting on it
Hope this helped you:)
