Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff
1 answer:
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Answer:
E = 1.04*10⁻¹ N/C
Explanation:
Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:
As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.
Replacing this values in the equation above, we can solve for a, as follows:
According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:
F = mp*a = q*E
For a proton, we have the following values:
mp = 1.67*10⁻²⁷ kg
q = e = 1.6*10⁻¹⁹ C
So, we can solve for E (in magnitude) , as follows:
⇒ E = 1.04*10⁻¹ N/C
Let
A = the amplitude of vibration
k = the spring constant
m = the mass of the object
The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω = the circular frequency.
The velocity is
v(t) = -ωA sin(ωt)
The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
Therefore
The KE (kinetic energy) is given by
The PE (potential energy) is given by
When the KE and PE are equal, then
For the oscillating spring,
Therefore
Answer:
A = the amplitude of vibration
k = the spring constant
m = the mass of the object
The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω = the circular frequency.
The velocity is
v(t) = -ωA sin(ωt)
The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
Therefore
The KE (kinetic energy) is given by
The PE (potential energy) is given by
When the KE and PE are equal, then
For the oscillating spring,
Therefore
Answer: