The whistle you use to call has a frequency of 21 kHz, so you do not recognize it also because your threshold of hearing is only 20 kHz. Since you are unable to hear sounds above this value, you request your friend to blow the whistle and move farther away. The frequency will decrease due to Doppler effect.
Answer:
The precision at which the number was measured
Explanation:
The number of digits that you use to record a measurement indicates the precision of the measurement.
Answer:
The electric charge, q (in coulomb units) = 5004 C
Given:
The charge stored as printed on NiMH battery, q = 1390 mAh
Solution:
To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:
1 mA = 
1 hour = 
Also, we know that the rate of flow of charge is electric current, I:
I = 
⇒ q = [tex]I\times t[tex] (1)
where
q = electric charge
I = current
t = time taken for flow of current
Using eqn(1), we get:
q = [tex]1390\times 10^{-3}\times 60\times 60[tex]
q = 5004 A-s = 5004 C
Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m
(2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp