Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz
velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
B = 0.018 T Ans,
Since, it is moving in a circular path, thus, centripetal force will act on it i.e.
F =
where, m is the mass of the object, v is the velocity and r is the radius of circular path.
And, since a positive charge is moving, it will create magnetic force which is equal to F = qvB
where q is the charge, v is the velocity of the particle and B is the magnetic field.
Now, the two forces will be equal,
i.e.
= qvB
⇒
= qB
⇒B =
<span>putting the values, we get,
</span>
use q = 1.6 * 10^ -19
⇒ B = 0.018 T
Answer:
a) Magnetic field strength, B = 2.397 * 10⁻⁷ T
b) Total energy density, U = 4.58 * 10⁻⁸ J/m³
c) Power flow per unit area, S = 13.71 W/m²
Explanation:
a) Electric field strength, E = 71.9 V/m
The relationship between the Electric field strength and the magnetic field strength in vacuum is:
E = Bc where c = 3.0 * 10⁸ m/s
71.9 = B * 3.0 * 10⁸
B = 71.9 / (3.0 * 10⁸)
B = 23.97 * 10⁻⁸
B = 2.397 * 10⁻⁷ T
b) Total Energy Density:
c)Power flow per unit area
Complete question is;
Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)
(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Answer:
A) P_out = 24 W
B) t = 1470 s
C) Q = 1140.72 KJ
Explanation:
We are given;
Input Power; P_in = 800 W
Efficiency; η = 3% = 0.03
A) Formula for efficiency is;
η = P_out/P_in
Making P_out the subject, we have;
P_out = η•P_in
P_out = 0.03 × 800
P_out = 24 W
B) We know that;
Power = work done/time taken
Thus;
P_out = mgh/t
We are given;
m = 3000 kg
h = 1.20 m
Thus, time is;
t = (3000 × 9.8 × 1.2)/24
t = 1470 s
C) amount of heat wasted is calculated from;
Q = (P_in - P_out)t
Q = (800 - 24) × 1470
Q = 1,140,720 J
Q = 1140.72 KJ
