Relativistic reversal of events. Figures 37-25 a and b show the (usual) situation in which a primed reference frame passes an un-primed reference frame, in the common positive direction of the x and x' axes, at a constant relative velocity of magnitude v . We are at rest in the unprimed frame; Bullwinkle, an astute student of relativity in spite of his cartoon upbringing, is at rest in the primed frame. The figures also indicate events A and B that occur at the following spacetime coordinates as measured in our unprimed frame and in Bullwinkle’s primed frame: Event Unprimed Primed A ( x A , t A ) ( x ' A , t ' A ) B ( x B , t B ) ( x ' B , t ' B ) In our frame, event A occurs before event B, with temporal separation ∆ t = t B − t A = 1.00 µ s and spatial separation ∆ x = x B − x A = 400 m. Let ∆ t' be the temporal separation of the events according to Bullwinkle. (a) Find an expression for ∆ t' in terms of the speed parameter ß ( = v/c ) and the given data. Graph ∆ t' versus ß for the following two ranges of ß: (b) 0 to 0.01 ( v is low, from 0 to 0.01 c ) (c) 0.1 to 1 ( v is high, from 0.1 c to the limit c ) (d) At what value of ß is ∆ t' = 0? For what range of ß is the sequence of events A and B according to Bullwinkle (e) the same as ours and (f) the reverse of ours? (g) Can event A cause event B , or vice versa? Explain. Figure 37-25 Problem 21, 22, 60, and 61.
Relativistic reversal of events. Figures 37-25 a and b show the (usual) situation in which a primed reference frame passes an un-primed reference frame, in the common positive direction of the x and x' axes, at a constant relative velocity of magnitude v . We are at rest in the unprimed frame; Bullwinkle, an astute student of relativity in spite of his cartoon upbringing, is at rest in the primed frame. The figures also indicate events A and B that occur at the following spacetime coordinates as measured in our unprimed frame and in Bullwinkle’s primed frame: Event Unprimed Primed A ( x A , t A ) ( x ' A , t ' A ) B ( x B , t B ) ( x ' B , t ' B ) In our frame, event A occurs before event B, with temporal separation ∆ t = t B − t A = 1.00 µ s and spatial separation ∆ x = x B − x A = 400 m. Let ∆ t' be the temporal separation of the events according to Bullwinkle. (a) Find an expression for ∆ t' in terms of the speed parameter ß ( = v/c ) and the given data. Graph ∆ t' versus ß for the following two ranges of ß: (b) 0 to 0.01 ( v is low, from 0 to 0.01 c ) (c) 0.1 to 1 ( v is high, from 0.1 c to the limit c ) (d) At what value of ß is ∆ t' = 0? For what range of ß is the sequence of events A and B according to Bullwinkle (e) the same as ours and (f) the reverse of ours? (g) Can event A cause event B , or vice versa? Explain. Figure 37-25 Problem 21, 22, 60, and 61.
Relativistic reversal of events. Figures 37-25a and b show the (usual) situation in which a primed reference frame passes an un-primed reference frame, in the common positive direction of the x and x' axes, at a constant relative velocity of magnitude v. We are at rest in the unprimed frame; Bullwinkle, an astute student of relativity in spite of his cartoon upbringing, is at rest in the primed frame. The figures also indicate events A and B that occur at the following spacetime coordinates as measured in our unprimed frame and in Bullwinkle’s primed frame:
Event
Unprimed
Primed
A
(xA, tA)
(
x
'
A
,
t
'
A)
B
(xB, tB)
(
x
'
B, t
'
B)
In our frame, event A occurs before event B, with temporal separation ∆t = tB− tA = 1.00 µs and spatial separation ∆x = xB− xA = 400 m. Let ∆t' be the temporal separation of the events according to Bullwinkle. (a) Find an expression for ∆t' in terms of the speed parameter ß(= v/c) and the given data. Graph ∆t' versus ß for the following two ranges of ß:
(b) 0 to 0.01 (v is low, from 0 to 0.01c)
(c) 0.1 to 1 (v is high, from 0.1c to the limit c)
(d) At what value of ß is ∆t' = 0? For what range of ß is the sequence of events A and B according to Bullwinkle (e) the same as ours and (f) the reverse of ours? (g) Can event A cause event B, or vice versa? Explain.
Find the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring).
d
Ag
= 2.51
dFe
×
Show that the units 1 v2/Q = 1 W, as implied by the equation P = V²/R.
Starting with the equation P = V²/R, we can get an expression for a watt in terms of voltage and resistance. The units for voltage, V, are equivalent to [?
v2
v2
A, are equivalent to J/C ✓ X . Therefore, 1
= 1
= 1 A V1 J/s
Ω
V-A X
= 1 W.
. The units for resistance, Q, are equivalent to ?
The units for current,
Please solve and answer the question correctly please. Thank you!!
Laboratory Experiments in Microbiology (12th Edition) (What's New in Microbiology)
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