A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
2 answers:
Answer:
F = 2.49 x 10⁻⁹ N
Explanation:
The electrostatic force between two charged bodies is given by Colomb's Law:
where,
F = Electrostatic Force = ?
k = colomb's constant = 9 x 10⁹ N.m²/C²
q₁ = charge on proton = 1.6 x 10⁻¹⁹ C
q₂ = second charge = 1.4 C
r = distace between charges = 0.9 m
Therefore,
<u>F = 2.49 x 10⁻⁹ N</u>
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Answer:
The points 2 and 4 should be connected.
Explanation:
To complete the circuit, we need to connect the two points which when connected, encompass the battery and the bulb in the circuit. The points 2 and 4 do the job, since they connect the terminal of the battery and the terminal of the bulb, and thus complete the circuit.
Therefore, the choice C is correct.
Refer to the diagram shown.
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)