Answer:21
Explanation:every body said
Answer:
I may be wrong
Explanation:
it it won't collapse because it is like a log logs don't sink when they are in water
The speed is changing its direction all the time. There
is an acceleration which changes the direction of the speed – that is called
centripetal acceleration. Only uniform linear motions are considered to have no
acceleration.
This is the general formula for acceleration
a = dv/dt
When calculating dv, you should keep in mind the change
in the velocity vector’s direction. You can easily see in a graph that with dt
tending to 0 (so the length of the arc covered is also tending to 0), the difference
between vectors Vf and V0 has a direction which is perpendicular to velocity
(the shorter the arc, the closest the angle is to 90 degrees).
There is a formula (which can be deducted from the
previous formula) which allows you to calculate the acceleration:
a = v^2/r
Let’s talk about the units:
v is in m/s
r is in m
so v^2/r
is in (m/s)^2/m = (m^2/s^2)/m = m/s^2
which is the same unit as dv/dt:
dv/dt = (m/s)/s= m/s^2
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for
DAC:

⇒ 
Now for the
BAC:

⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle,
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 

After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x = 
I am pretty sure it is A Becoming warm
Since it’s moving and causing friction which makes it warm
Hope this helps
Mark me brainliest