A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll
2 answers:
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Answer:
C) The rate of change from potential to kinetic energy is exponential.
Explanation:
As we know that total mechanical energy must be conserved here
As we know that there is no friction force on this system of electromagnet and nail
So here we can say that that
Magnetic potential energy of the nail + electromagnet system will convert into kinetic energy of the nail as it is released.
So we will have
Initial potential energy = work done to move it away by 10 cm
now as the nail is released then this potential energy will start to convert into kinetic energy
So here correct answer must be
C) The rate of change from potential to kinetic energy is exponential.
Answer:
same value in R and 2R E = E₀ = σ / 2ε₀
Explanation:
For this exercise we use Gauss's law
Ф = E. dA = /ε₀
We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero
As the sheet has two faces
2E A = q_{int} /ε₀
The charge inside the cylinder is
σ = q_{int} / A
q_{int} = σ A
We substitute
E = σ / 2ε₀
We see that this expression is independent of the distance, so it has the same value in R and 2R
E = E₀ = σ / 2ε₀
Answer:
A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)
Explanation:
A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.
Lens 1. More to the left
let's use the constructor equation
where f is the focal length, p and q are the distance to the object and the image, respectively,
We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.
1 / q₁ = 0.04756
q₁ = 21.0227 cm
this image is the object for the second lens that has f₂ = 14.8 cm
the distance must be measured from the second lens
p₂ = 39.4 -q₁
p₂ = 39.4 -21.0227
p₂ = 18.38 cm
let's use the constructor equation
1 / q₂ = 1 / f - 1 / p2
= 0.01316
q₂ = 75.98 cm
measured from the second lens
B) the position of the final image with respect to the first lens is
q₂’= q₂ + 39.4
q₂'= 75.98 +39.4
q₂' = 115.38 cm
C) the magnification of a lens is
m = - q / p
in this case the image measured from lens 2 is q2 = 75.98 cm
the distance to the object from the first lens is p1 = 50cm
m = - 75.98 / 50
m = -1.5 X
the negative sign indicates that the image is inverted