Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:

Explanation:
Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.
During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.
Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.
Answer:
non linear square relationship
Explanation:
formula for centripetal force is given as
a = mv^2/r
here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have
a = constant × v^2
a α v^2
hence non linear square relationship
Answer: MR²
is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center
Explanation:
Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.
I = ∫ r² dm
= R²∫ dm
MR²
where M is total mass and R is radius of the hoop .
Answer:
For vector u, x component = 10.558 and y component =12.808
unit vector = 0.636 i+ 0.7716 j
For vector v, x component = 23.6316 and y component = -6.464
unit vector = 0.9645 i-0.2638 j
Explanation:
Let the vector u has magnitude 16.6
u makes an angle of 50.5° from x axis
So 
Vertical component 
So vector u will be u = 10.558 i+12.808 j
Unit vector 
Now in second case let vector v has a magnitude of 24.5
Making an angle with -15.3° from x axis
So horizontal component 
Vertical component 
So vector v will be 23.6316 i - 6.464 j
Unit vector of v 