To solve this problem, we need to use the principles of electric and kinematic physics.
1. First, let's calculate the force exerted on the particle due to the electric field. We can use the equation F = qE, where F is the force, q is the charge, and E is the electric field strength.
Given:
Charge (q) = -4.0 mC (the negative sign indicates that it is an electron)
Electric field (E) = 60 N/C (positive y-direction)
F = (-4.0 x 10^(-3) C) * (60 N/C) [Conversion from milliCoulombs to Coulombs]
F = -0.24 N (negative sign indicates the force is in the opposite direction of the electric field)
2. Since the electric field is perpendicular to the initial velocity of the particle, no work is done on the particle, and its kinetic energy remains constant. Thus, mechanical energy is conserved.
3. The kinetic energy (KE) of a moving object with mass (m) and velocity (v) is given by the equation KE = 0.5 * m * v^2.
Given:
Mass (m) = 50 g (conversion to kg: 0.05 kg)
Initial velocity (v) = 25 m/s
KE1 = 0.5 * (0.05 kg) * (25 m/s)^2
KE1 = 0.5 * (0.05 kg) * (625 m^2/s^2)
KE1 = 7.8125 J
4. Since the mechanical energy is conserved, the final kinetic energy (KE2) of the particle is also equal to 7.8125 J.
5. The final kinetic energy (KE2) can be related to the final velocity (vf) using the equation KE = 0.5 * m * vf^2.
7.8125 J = 0.5 * (0.05 kg) * vf^2
7.8125 J = 0.025 kg * vf^2
vf^2 = 7.8125 J / 0.025 kg
vf^2 = 156.25 m^2/s^2
vf = √(156.25) m/s
vf = 12.5 m/s
Therefore, the speed of the particle 5.0 s after it enters this region is 12.5 m/s.
show your work if the answer to this question is 35m\s
A particle (q-4.0 mC.m-50 g) has a velocity of 25 m/s in the positive x direction when it first enters a region where the electric field is uniform (60 N/C in the positive y direction). What is the speed of the particle 5.0 s after it enters this region?
is simple steps
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