The surface of an incline is coated with an experimental substance that is intended to reduce the frictional force between a blo
ck and the
surface of the incline. A 2 kg block is placed at the top of the incline at a height of 1.8 m, as shown in the figure. After the block is released
from rest, the block slides down the incline and a motion detector at the bottom of the incline measures the block's speed as 5.8 after the
block is no longer on the incline. Which of the following claims is correct about the experimental substance?
The experimental substance reduced all the frictional force because all the gravitational potential energy of the Earth-block
system at the top of the incline was converted into the kinetic energy of the block at the bottom of the incline.
The experimental substance did not reduce all the frictional force because some of the gravitational potential energy of the
Earth-block system at the top of the incline was converted into nonmechanical energy
с
The effectiveness of the experimental substance cannot be determined because the speed of the block at the bottom of the
inline, as measured by the motion detector, indicates that the block has more energy at this location than the Earth-block
system had at the top of the incline.
D
The effectiveness of the experimental substance cannot be determined without knowing the magnitude of the frictional
force between the block and the incline before and after the experimental substance was applied to the incline.
Lo
surface of the incline. A 2 kg block is placed at the top of the incline at a height of 1.8 m, as shown in the figure. After the block is released
from rest, the block slides down the incline and a motion detector at the bottom of the incline measures the block's speed as 5.8 after the
block is no longer on the incline. Which of the following claims is correct about the experimental substance?
The experimental substance reduced all the frictional force because all the gravitational potential energy of the Earth-block
system at the top of the incline was converted into the kinetic energy of the block at the bottom of the incline.
The experimental substance did not reduce all the frictional force because some of the gravitational potential energy of the
Earth-block system at the top of the incline was converted into nonmechanical energy
с
The effectiveness of the experimental substance cannot be determined because the speed of the block at the bottom of the
inline, as measured by the motion detector, indicates that the block has more energy at this location than the Earth-block
system had at the top of the incline.
D
The effectiveness of the experimental substance cannot be determined without knowing the magnitude of the frictional
force between the block and the incline before and after the experimental substance was applied to the incline.
Lo
2 answers:
You might be interested in
1) Electric potential inside the sphere:
2) Ratio Vcenter/Vsurface: 3/2
3) Find graph in attachment
Explanation:
1)
The electric field inside the sphere is given by
where
is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
r is the distance from the centre at which we compute the field
For a radial field,
Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,
The potential at the surface, V(R), is that of a point charge, so
Therefore we can find the potential inside the sphere, V(r):
2)
At the center,
r = 0
Therefore the potential at the center of the sphere is:
On the other hand, the potential at the surface is
Therefore, the ratio V(center)/V(surface) is:
3)
The graph of V versus r can be found in attachment.
We observe the following:
- At r = 0, the value of the potential is , as found in part b) (where
)
- Between r and R, the potential decreases as
- Then at r = R, the potential is
- Between r = R and r = 3R, the potential decreases as , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (
)
Learn more about electric fields and potential:
#LearnwithBrainly
Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here
is linear charge density and r is distance of the point where we have to find the electric field
So
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m