Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
Answer:
Acceleration will increase.
Explanation:
The relation between force, mass and acceleration according to the Newton's second law of motion is given as:
F = ma
We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:
Since, F is constant, the acceleration of the truck is inversely proportional to the mass.
There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.
Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.
So, the acceleration of the truck will increase.
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
Every person is different. But for a planet-wide overall average that roughly represents all human beings on Earth, the figures usually used are:
from 20 Hz to 20,000 Hz .
Answer:
<em>1.01 W/m</em>
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.
area of the pipe per unit length A = 
= 
m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>
