value is identified in Chapter 22). What is the ratio of the re... | Filo (2024)

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Electrons in a conductor have no motion in the absence of a potential difference across it.

The radius of curvature of the face of a plano-convex lens is 12cm and its refractive index is 1.5.(a) Find the focal length of the lens. The plane surface of the lens is now silvered(b) At what distance from the lens will parallel rays incident on the convex face converge?(c) Sketch the ray diagram to locate the image, when a point object is placed on the axis, 20cm from the lens.(d) Calculate the image distance when the object is placed as in part (c).

A metal wire of resistance 3Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60∘ at the centre, the equivalent resistance between these two points will be(a) 27​Ω(b) 25​Ω(c) 512​Ω(d) 35​Ω

Estimate an expression for the free fall time of the Sun if it were to disappear one day. Use mass, the gravitational constant, and the Sun's radius as variables.

PLEASE READ THE QUESTION CAREFULLY!! I WILL FOR SURE GIVE YOU THUMBS UP!! (Guaranteed)Chegg solution citation:https://www.chegg.com/homework-help/4-kg-block-ais-dropped-height-800-mm-onto-9-kg-block-bwhich-chapter-19-problem-134p-solution-9780077275556-excMy question is about the solution, so PLEASE read carefully!!!!!!!!!!A 4-kg block A is dropped from a height of 800 mm onto a 9-kg block B which is at rest. Block B is supported by a spring of constant k = 1500 N/m and is attached to a dashpot of damping coefficient c = 230 N s/m. Knowing that there is no rebound, determine the maximum distance the blocks will move after the impact.Solution: Given data: The mass of block A, m = 4 kg The mass of block B, m = 9 kgThe constant of the spring, k = 1500 N/mThe damping coefficient of the dashpot, C = 230 N-s/mFig. P19.134Block A is dropped from a height of 0.8 mA does not rebound after collision with B.Consider block A just before impact.Apply the law of conservation of momentum for blocksThe static deflection caused by block A,The velocity, v = 2ghA and B before and just after collision. Then,-mgX = k(4)(9.81) 1500mv + mgV = m + mv= /(2)(9.81 m/s^2)(0.8 m)(9)(3.962) + 0 = (4 + 9)v= 3.962 m/sv = 1.219 m/s The common velocity of A and B just after collision is v = 1.219 m/s-0.02619 mx = 0.02619 m We can write,MY QUESTION: refer to the blue boxx = v= 1.219 m/s When t = 0, x = -0.02619 mSo here, the solution is saying that the initial positional condition is at the static deflection.... how is this possible? So right now my intuition is that when A goes on to B block, that is when t = 0 starts "counting the time".... adding on, when we say "deflection" it means there is a displacement, that means the block has moved from (xo = 0 m) to (xo = 0.02619 m) in 0 seconds???? Please Help!! I will give you thumbs up!!!! please,,,.Please Help..... the Solution Chegg Citation is the below URL

A Cassegrain astronomical telescope uses two mirrors to form the image. The larger (concave) objective mirror has a focal length f 1 = +50.0 cm. A small convex secondary mirror is mounted 43.0 cm in front of the primary. As shown in Figure 27-31 light is reflected from the secondary through a hole in the center of the primary, thereby forming a real image 8.00 cm behind the primary mirror. What is the radius of curvature of the secondary mirror?