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
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.
To solve this problem, we need to use the given information and apply the relevant principles of physics. Here are the step-by-step calculations:
Step 1: Convert the charge from microcoulombs (µC) to coulombs (C):
Given charge, q = 4.0 mC = 4.0 × 10^(-3) C.
Step 2: Calculate the force on the particle due to the electric field:
Given electric field, E = 60 N/C.
Electric force, F = qE = (4.0 × 10^(-3) C) × (60 N/C).
Step 3: Determine the acceleration of the particle using Newton's second law:
Given mass, m = 50 g = 50 × 10^(-3) kg.
Acceleration, a = F/m = [(4.0 × 10^(-3) C) × (60 N/C)] / (50 × 10^(-3) kg).
Step 4: Find the change in velocity after 5.0 s:
Given initial velocity, u = 25 m/s.
Time, t = 5.0 s.
Final velocity, v = u + at.
Step 5: Calculate the speed of the particle after 5.0 s:
Speed is the magnitude of velocity, so we need to determine the magnitude of v.
Speed, v = |v| = |u + at|.
Step 6: Substitute the values into the formula and calculate the speed:
v = |25 m/s + [(4.0 × 10^(-3) C) × (60 N/C)/(50 × 10^(-3) kg)] × 5.0 s|
Step 7: Simplify and solve for the speed:
v = |25 m/s + [(4.0 × 10^(-3) C) × 60 N/C × 5.0 s/(50 × 10^(-3) kg)]|
Step 8: Perform the calculations:
v = |25 m/s + 4 C × 60 N/C × 5.0 s/(50 kg)|
Step 9: Simplify the expression inside the magnitude brackets:
v = |25 m/s + 4 × 60 N × 5.0 s/50 kg|
Step 10: Perform the remaining calculations and find the magnitude of v:
v = |25 m/s + 4 × 60 N × 5.0 s/50 kg|
v = |25 m/s + 24 N × 5.0 s/50 kg|
v = |25 m/s + 24 N × 5.0 s/50 kg|
v = |25 m/s + 24 × 5.0 N s/50 kg|
v = |25 m/s + 120 N s/50 kg|
v = |25 m/s + 2.4 N s/kg|
v = |25 m/s + 2.4 m/s|
Step 11: Find the magnitude of v by summing the two speeds:
v = 25 m/s + 2.4 m/s
v = 27.4 m/s
Therefore, the speed of the particle 5.0 s after it enters the region is 27.4 m/s.
To find the speed of the particle 5.0 seconds after it enters the region, we can follow these steps:
Step 1: Calculate the force due to the electric field on the charged particle.
The force (F) on a charged particle in an electric field (E) is determined by the equation F = q * E, where q is the charge of the particle. In this case, the particle has a charge of -4.0 mC (negative because it's an electron), and the electric field is 60 N/C in the positive y direction. Therefore, the force on the particle is F = (-4.0 mC) * (60 N/C).
Step 2: Determine the acceleration of the particle.
Newton's second law states that the force acting on an object is equal to its mass (m) multiplied by its acceleration (a). Rearranging this equation, we can solve for acceleration: a = F / m. In this case, the mass of the particle is 50 g, which is 0.05 kg. Thus, the acceleration of the particle is a = (force calculated in step 1) / (0.05 kg).
Step 3: Calculate the change in velocity over 5 seconds.
Since the acceleration of the particle is constant, we can use the equation v = u + a * t to find the change in velocity (v) over a given time (t), where u is the initial velocity. In this case, the initial velocity of the particle is 25 m/s in the positive x direction. Plugging in the values, we get v = 25 m/s + (acceleration calculated in step 2) * 5 s.
Step 4: Determine the speed of the particle.
The speed of the particle is the magnitude of its velocity. Since the velocity is in the positive x direction, the magnitude is just the absolute value. Thus, the speed of the particle is |velocity calculated in step 3|.
By following these steps and plugging in the given values, you can find the speed of the particle 5.0 seconds after it enters the region.