Elastic potential energy is given by formula
here we know that
Now using above formula we have
So elastic potential energy in the chord is 14400 J
Answer:
2954.6 N/C, 46.36 degree from positive axis
Explanation:
E1 = 1300 N/C, θ1 = 35 degree
E2 = 1700 N/C, θ2 = 55 degree
Now write the electric fields in vector form
E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j
E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j
Resultant electric field
E = E1 + E2
E = 1064.9 i + 745.6 j + 975.08 i + 1392.6 j
E = 2039.08 i + 2138.2 j
Magnitude of E
E = sqrt (2039.08^2 + 2138.2^2)
E = 2954.6 N/C
Let it makes an angle Φ from X axis
tan Φ = 2138.2 / 2039.08 = 1.049
Φ = 46.36 degree from positive X axis.
Answer:
C
Explanation:
During an investigation you want to investigate different variables.
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
