We have that the sea level pressure for Leh area is 1150mb mathematically given as
Ps= 1150 mb
<h3>
Sea level pressure</h3>
Question Parameters:
Ladakh is 800 mb.
<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m
increase in height with respect to sea level corresponds to 10 mb pressure,
Generally, for 3500m the pressure change will be 350 mb.
Therefore, here for the sea level <em>pressure</em> we need to add,
Ps=800+350
Ps= 1150 mb
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B. velocity at position x, velocity at position x=0, position x, and the original position
In the equation
=
+2 a x (x - x₀)
= velocity at position "x"
= velocity at position "x = 0 "
x = final position
= initial position of the object at the start of the motion
The radius of the sphere in meters is ,r =
Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,
r/10 = tan [arctan(1/2)] = tan (1/2)
Since, tan (θ/2) = 1-cos(θ) / sin(θ)
we find that,
r/10 = 
Hence, r = 
So, the radius of the sphere in meters is ,r =
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Each hour 430 quintillion Joules of energy from the sun hits the Earth.
In a year it is very hard to determine because of the night and different light levels.
Answer:
a) 
b) 
c) 
Explanation:
the gravitational potential energy is given by:


The kinetic energy is given by:

the initial kinetic energy is zero because the motion started from rest, so:

applying the conservation of energy theorem:

The work done by the friction force is given by:

the angle of the force is 180 degrees because it's against the movement:
