Superluminal jets. Figure 37-29 a shows the path taken by a knot in a jet of ionized gas that has been expelled from a galaxy. The knot travels at constant velocity v → at angle θ from the direction of Earth. The knot occasionally emits a burst of light, which is eventually detected on Earth. Two bursts are indicated in Fig. 37-29 a, separated by time t as measured in a stationary frame near the bursts. The bursts are shown in Fig. 37-29 b as if they were photographed on the same piece of film, first when light from burst 1 arrived on Earth and then later when light from burst 2 arrived. The apparent distance D app traveled by the knot between the two bursts is the distance across an Earth-observer’s view of the knot’s path. The apparent time T app between the bursts is the difference in the arrival times of the light from them. The apparent speed of the knot is then V app = D app /T app . In terms of v , t, and θ, what are (a) D app and (b) T app ? (c) Evaluate V app for v = 0.980 c and θ = 30.0°. When superluminal (faster than light) jets were first observed, they seemed to defy special relativity—at least until the correct geometry (Fig. 37-29 a ) was understood. Figure 37-29 Problem 63.
Superluminal jets. Figure 37-29 a shows the path taken by a knot in a jet of ionized gas that has been expelled from a galaxy. The knot travels at constant velocity v → at angle θ from the direction of Earth. The knot occasionally emits a burst of light, which is eventually detected on Earth. Two bursts are indicated in Fig. 37-29 a, separated by time t as measured in a stationary frame near the bursts. The bursts are shown in Fig. 37-29 b as if they were photographed on the same piece of film, first when light from burst 1 arrived on Earth and then later when light from burst 2 arrived. The apparent distance D app traveled by the knot between the two bursts is the distance across an Earth-observer’s view of the knot’s path. The apparent time T app between the bursts is the difference in the arrival times of the light from them. The apparent speed of the knot is then V app = D app /T app . In terms of v , t, and θ, what are (a) D app and (b) T app ? (c) Evaluate V app for v = 0.980 c and θ = 30.0°. When superluminal (faster than light) jets were first observed, they seemed to defy special relativity—at least until the correct geometry (Fig. 37-29 a ) was understood. Figure 37-29 Problem 63.
Superluminal jets. Figure 37-29a shows the path taken by a knot in a jet of ionized gas that has been expelled from a galaxy. The knot travels at constant velocity
v
→
at angle θ from the direction of Earth. The knot occasionally emits a burst of light, which is eventually detected on Earth. Two bursts are indicated in Fig. 37-29a, separated by time t as measured in a stationary frame near the bursts. The bursts are shown in Fig. 37-29b as if they were photographed on the same piece of film, first when light from burst 1 arrived on Earth and then later when light from burst 2 arrived. The apparent distance Dapp traveled by the knot between the two bursts is the distance across an Earth-observer’s view of the knot’s path. The apparent time Tapp between the bursts is the difference in the arrival times of the light from them. The apparent speed of the knot is then Vapp = Dapp/Tapp. In terms of v, t, and θ, what are (a) Dapp and (b) Tapp? (c) Evaluate Vapp for v = 0.980c and θ = 30.0°. When superluminal (faster than light) jets were first observed, they seemed to defy special relativity—at least until the correct geometry (Fig. 37-29a) was understood.
SARET CRKS AUTOWAY
12. A stone is dropped from the top of a cliff. It is seen to hit the ground below
after 3.55 s. How high is the cliff?
13. A ball is dropped from rest at the top of a building that is 320 m tall. Assuming
no air resistance, what is the speed of the ball just before it strikes the ground?
14. Estimate (a) how long it took King Kong to fall straight down from the top
of the Empire State Building (280m high), and (b) his velocity just before
"landing".
Useful equations
For Constant Velocity:
V =>
D
X = V₁t + Xo
For Constant Acceleration:
Vr = V + at
X = Xo+Vot +
v=V+2a(X-Xo)
\prom = V +V
V velocity
t = time
D Distance
X = Final Position
Xo Initial Position
V = Final Velocity
Vo Initial Velocity
a = acceleration
For free fall
Yf
= Final Position
Yo Initial Position
g = 9.80
m
$2
For free fall:
V = V + gt
Y=Yo+Vo t +
+gt
V,² = V₁²+2g (Y-Yo)
V+Vo
Vprom=
2
6
Solve the problems
A 11 kg weight is attached to a spring with constant k = 99 N/m and subjected to an external force
F(t) =-704 sin(5t). The weight is initially displaced 4 meters above equilibrium and given an
upward velocity of 5 m/s. Find its displacement for t> 0.
y(t)
ון
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