The answer is D. As the ambulance gets closer, the sound waves are compressed relative to the person; so the frequency increases.
Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
= 
So the lines terminating at - 2 micro coulomb
= 
So, the number of filed lines terminates at - 2 micro Coulomb are 6.
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
Speed = (distance covered) / (time to cover the distance).
The speed of anything that covers 42 meters in 7 seconds is
(42 meters) / (7.0 seconds)
= (42 / 7.0) (meters/second)
= 6.0 m/s .
