Normal reaction force on the block while it is at rest on the inclined plane is given as

here we know that
m = 46 kg

now we will have

now the limiting friction or maximum value of static friction on the block will be given as


Above value is the maximum value of force at which block will not slide
Now the weight of the block which is parallel to inclined plane is given as

here we know that

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.
So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>
Volumetric cylinders and volumetric flasks
Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
Thermal energy.
The more heat an object has, the faster the particles will move and bump into each other.