Answer:
a)
two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)
b)
both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Explanation:
Given that;
L = 0.26 m
k = 180 N/m
x = 0.039 m
a)
we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.
b)
Spring force F = kx
F = 180 × 0.039
F = 7.02 N
Now, Electrostatic force F = Keq²/r²
where r = L + x = ( 0.26 + 0.039 )
we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2
so from the equation; F = Keq²/r²
Fr² = Keq²
q = √ ( Fr² / Ke )
we substitute
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ (0.627595 / 9×10⁹)
q = √(6.97 × 10⁻¹¹)
q = 8.35 × 10⁻⁶ C
Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Answer:
19.2m/s
Explanation:
Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,
plug in known varibles,
v=2.4*8
v=19.2m/s
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ = 
A₂ = Area at the throat
A₂ = 
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Answer:
0.04
Explanation:
Fraction of power converted to sound = 80% = 0.08
Fraction incident upon each eardrum onstage = 0.08/2 = 0.04
<span>Radio waves just like light waves can be reflected refracted and diffracted and polarized. The answer is True. </span>These characteristics are the common phenomena for electromagnetic (EM) waves, and Radio Waves are electromagnetic Waves so much so that they obey reflection, refraction, and diffraction.
