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s = 4(0) 2
v = ds dt = 8tms
− 1
a = d
(^2) s dt^2 = 8ms
− 2
50 km/hr = 50 × 103 3600 = 13.^9 ms
− 1
t = v a = 13.98 = 1.^7 s
x = x 0 + v 0 t + 1 2 at
2
= 1 + 0 × 8 + 12 × 8 × 1. 72 = 13m
ac = v
2 r
= 13.^9
2 30 = 6. 4 ms−^2 towards the center of the curve
y = 1 2 ay^ t
2
⇒ t =
2 y g =
2 × 1. 4
- 8 = 0.^53 s
x = vx 0 t = 13. 9 × 0 .53 = 7. 4 m
vy = ay t = 9. 8 × 0 .53 = 5. 2 ms − 1
v =
v x^2 + v^2 y =
- 92 + 5. 22 = 15ms−^2
To work out direction need to calculate θ
tan θ =
vy vx =^
- 2
- 9
θ = 20. 5 o below horizontal
P =
F
A
Pf = Pi +
mg
A
= 1. 17 × 10
5
5 × 9. 8
12. 0 × 10 −^4
= 1. 58 × 10
5 P a
quickly ⇒ adiabatic P V γ = const
γ =
CP
CV
R +
1 2 f R 1 2 f R^
PiV
5 / 3 i =^ Pf^ V^
5 / 3 f
- 17 × 10 5 × (0. 20 × 12. 00 × 10 − 4 ) 5 / 3 = 1. 58 × 10 5 × (hf × 12 × 10 − 4 ) 5 / 3
(hf × 12 × 10 − 4 ) 5 / 3 = 8. 8635 × 10 − 7
hf × 12 × 10 − 4 = 2. 004 × 10 − 4
hf = 16. 7 cm
P V = const PiVi = Pf Vf
- 17 × 10 5 × 0. 20 × 12. 00 × 10 − 4 = 1. 58 × 10 5 × hf × 12 × 10 − 4
hf = 14. 8 cm
P = kA|
dT
dx
P is constant along the rod
⇒ P = 385 × 10 × 10
− 4 ×
= 9. 63 W
9 .625 = 50. 2 × 10. 0 × 10
− 4 ×
L 2
L 2 = 0. 156 m
∆Eint = 0 as isothermal
∆Eint = Q + W W = 0 as volume does not change ∆Eint = − 233 kJ
WC→A = ∆EintC→A − QC→A QC→A = 0 a adiabatic ∆EintC→A = −(∆EintA→B + ∆EintB→C ) as no change in internal energy over cycle = −(0 − 233) = 233kJ WC→A = 233kJ
Alternatively could use W = −
P dV and integrate the expression.
This takes longer.
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S2P " S 1 P = 13 " 12 = 1 m
thus # = 1/2 = 0.5 m
v = #f
f =
v
= 700 Hz
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(S 2 P + s 2 ) " (S 1 P + s 1 ) = 0.75 eqn(1)
from Pythagoras :
(S2P + s2) 2 = 25 + (S 1 P + s 1 ) 2 eqn(2)
(S2P + s2) = 12.75 + s
Substitute into eqn(2)
(12.75 + s1)
2 = (12 + s1)
2
which can be solved for s
The answer is then 12 + s
First find s1:
s 1 =
= 4.29 # 4.3 m
new distance from S2 : 13 + s2 = 12.75 + 4.
new distance from S2 = 17 m
(c) Have to do (ii) before (i) to find the train speed.
(ii) If the frequency appears lower to the observer, the source and the observer must be moving away from each other: their velocities will be negative
f' =
v + vo
v " vs
(f
340 + x
x =
) 330 " 340 = 14.8 * 15 m/s
(i) Train moving away from stationary observer:
f' =
v
v + vs
'f^ =^
' 600 = 575 Hz
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(iii) Treat the wall as stationary observer: therefore f = 575 Hz
The cyclist is moving towards the wall, which has now become a source, reflecting sound at 575 Hz.
Frequency received by the cyclist moving towards the wall:
f' =
v + vo
v
'f^ =^
' (^575 =^ 591.9^ )^592 Hz
Beating frequency: 592 – 558 = 34 Hz
