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Homework Statement I am seeking a derivation of the formula for greatest detail or maximum resolution of an astronomical telescope, which is: Homework Equations M = fo/foe where: M: magnification fo: focal length of objective lens foe: distance of primary image from the eyepiece...

Comments anyone? Help and advice gratefully received... :)

I am not using 40pi * 1200: the rotational speed is 40pi which is used in the expression which requires angular velocity in rad/s. The 40pi comes from converting 1200rpm to rad/s. I then use the value 40pi for w (omega) in the expression..l.

Homework Statement A rectangular coil of 200 turns has a length of 200 mm and width 120 mm. The coil rotates with a constant angular speed of 1200 revolutions per minute about an axis through the midpoints of its longer sides in a uniform magnetic field of 2.4 x 10-2 T. Starting from a...

Homework Statement The speed of sound in a metal rod is 3600 m s -1. The rod is 1.20m long and clamped at one of its ends. (a) Determine the frequency of its vibration if longitudinal waves are established in the rod and it is vibrating in its first overtone mode. (d) Determine the...

Quite so... many thanks for your help!

Homework Statement A ray of monochromatic yellow light is incident in air on an equilateral triangular glass prism. This ray is in the same plane as the equilateral triangular cross section of the prism; the angle between the ray and the prism face is 60°, and the refractive index of the...

I guess it's possible the larger sphere could become a dipole if the smaller sphere was small enough and hence had a much higher charge density and hence electric field. Is this correct?

Homework Statement Two insulated metal spheres of equal sizes are given equal positive charges. The two spheres are brought very close to each other without touching. Draw the new charge distribution on the spheres if one sphere is much larger than the other. Homework Equations...

Homework Statement Two wooden blocks of mass 8 kg and 4 kg respectively approach each other from opposite directions on a smooth level surface at a relative speed of 16 m s^1 . After a head-on collision they separate at a relative speed of 6 m s^1 . The initial velocity of the 8 kg block...

Thanks - point taken - I reworked the expressions for Fy and Fx in terms of string tension: sumFy = FTcos 10 - mg cos 20 sumFx=FT sin10 - mg sin 20 m x ax = mg cos 20. tan 10 -mg sin 20 ax =g(cos20.tan10 -sin 20) N.B. ax is x acceleration with xy co-ordinates of the slope...

Homework Statement A bus is descending a uniform 20 degree slope. It brakes with constant deceleration. A pendulum moves 10 degrees away from the vertical to the downward side. Find the acceleration of the bus. ....| .../| ... ../.| .../10| .../...|.../ ...O...|.../ ....|/...

Homework Statement A rock is thrown upwards at an angle of 35.0° to the horizontal. The rock hits a signpost 15.0 m away at a point 2.00 m above the level from which it was thrown. Calculate the initial velocity of the rock. Homework Equations d = v * t d= displacement...

After further thought it was clear that your explanation was only a trivial step short of the answer. However, I would be grateful if you would please check that my reasoning is correct: (a) Wind direction - the expression 8x + 16*cos30y (x,y instead of i,j) points to the co-ordinates (8...

Once again - thank you. I'm still puzzled as to how solve for two unknowns in the expression for the wind direction... the Wind is directed as = 8 i + 16*cos30 j

Please forgive my unfamiliarity with imaginary numbers. Do the i's and j's also represent the y and x co-ordinates? I am unsure how to solve the expression to find the co-ordinates.

Thanks for the hint. What is true is that WG (wind speed relative to the ground) must be the same for both cyclists. Following your suggestion I drew separate vector diagrams for the two cyclists: ...^.^ ...WG.../...| .../...|WX WG = WX +XG .../.....| -------------> XG...

I wonder if taking Y's direction as an absolute value affects the problem - however I feel that if a compass bearing is quoted then it should be treated as a compass bearing - still puzzled by the word relative in "relative to X".

Homework Statement To a cyclist X traveling at 8 km h-1 due east the wind appears to blow from the south. Another cyclist Y travels at 16 km h-1 at N30°W relative to X. The direction of the wind as experienced by Y is from the west. Calculate the speed and direction of the wind...

Please check my working - I don't get the specified answer - but it's possible that answer counted the storeys from the ground up... I now see my mistake in squaring one side of the expression... starting from there: 0.1^2+8n/g+(0.2 x SQRT(8n/g)) = (8n+8)/g Multiply both sides by g...

Still puzzled... n should give the number of storeys A is from the top... Following your suggestion, I have: 0.12+ (8n)/g = (8n+8)/g then 0.01g + 8n = 8n +8 The n's still cancel - I've obviously done something wrong The expected answer is 6

Homework Statement An object falls from the top of a 103.6m high building. The vertical distance between two consecutive windows A and B is 4.0m. The first window/storey is also 4m from the top of the building. The object reaches window B 0.1s after it has reached window A. Which...

Thank you for your reassurance. I lacked the confidence to be certain the second vector diagram was the equivalent of the first - especially because vector arithmetic is new to me. I would like to ask this naive question - if a, b and c are vector quantities and they are related: a = b+c then...

Homework Statement A car is traveling at 12ms^-1. To the passenger in the car the wind appears to be blowing at 8.0ms^-1 at right angles to the road. What is the magnitude and direction of the velocity of the wind with respect to the ground. Homework Equations I can think of two...

Thank you to everyone who gave their time to help me with this problem. It has served to identify the areas I must work on.

Many thanks for the help during the "holidays"... Because d = 1/2 * g * (2n-1) then n = (2*d/g + 1) / 2 = d/g + 1/2 However, when n is substituted into D = 1/2*g*n^2 as D = 1/2*g*(d/g + 1/2)^2 it does not produce the expected result. I guess I have a problem with...

Thanks for the hint - clearly the distance d is given by d = 1/2gn^2 - 1/2g(n-1)^2 which (according to my rusty maths) simplifies = 1/2g(2n-1) It appears that t^2 = 2d/g is useful here but substitution appears to give nonsense. I 've obviously missed or misunderstood something.

Homework Statement A body falls vertically from rest. During the nth second it falls a distance d. Prove that by the end of the nth second it has fallen a total distance (D) of (2d+g)^2/8g Homework Equations x = x0 + V0 + 1/2at^2 where x0 = initial position v0 = initial velocity a...