Missing an attached sheet with the sound waves. Please add the attached sheet so I can properly help.
Thanks! :D
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Answer:
Add Ff from Fa
Explanation:
Fnet = sum of all force
horizontal net force = Ff + Fa
Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
The approximate radius of the nucleus of this atom is 4.656 fermi.
b) 
k=
= Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
Force of repulsion between two protons on opposite sides of the diameter
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
<u>Given that:</u>
Ball dropped from a bridge at the rate of 3 seconds
Determine the height of fall (S) = ?
As we know that, S = ut + 1/2 ×a.t²
u =initial velocity = 0
a= g =9.81 m/s (since free fall)
S = 0+ 1/2 × 9.81 × 3²
<em> S = 44.145 m</em>
<em>44.145 m far is the bridge from water</em>
