Riders on the Tower of Doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.5
1 answer:
Newton's second law and kinematics allow finding the answers for the changes in weight when the body stops are:
A) The apparent weight is 3548 N directed upwards
B) The factor of change of weights is 5
Given parameters
- Free fall time t1 = 2.0 s
- The time tot stops at t2 = 0.5 s
- Mass m = 65 kg
To find
A) Apparent weight
B) Factor that exceeds the actual weight
Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.
This exercise we will do them in two parts
1st part. Freefall
We look for the velocity at the end of the trajectory, as part of rest the initial velocity is zero
v = v₀ - g t
Where v is the velocity, vo the initial velocity, g the acceleration due to gravity and y time
v = -gt
v = -9.8 2.0
v = -19.6 m / s
The negative sign indicates that the velocity is directed downwards
2nd part. Braking movement
We look for the acceleration to stop the body, in this case the final velocity is zero
= v + a t
0 = v + at
a = - v / t
a =
a = + 39.2 m / s²
The positive sign indicates that the acceleration is directed upward.
A) Newton's second law states that the force is proportional to the mass and the acceleration of the body
∑ F = m a
F -W = m a
F = m (a + g)
F = 65 (39.2 + 9.8)
F = 3185 N
Body weight is
W = mg
W = 65 9.8
W = 637 N
therefore the apparent weight is
ΔW = F -W
ΔW = 3185 - 637
Δw = 2548 N
B) the weight change factor
Factor = F / W
Factor =
factor = 5
In conclusion, using Newton's second law and kinematics, we can find the answers for the changes in weight when the body stops are:
A) The apparent weight is 3548 N directed upwards
B) The factor of change of weights is 5
Learn more about Newton's second law and kinematics here:
You might be interested in
Answer:
The magnitude of the magnetic field vector is 1.91T and is directed towards the east.
The steps to the solution can be found in the attachment below.
Explanation:
For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.
Fm = Fg
qvBSin(theta) = mg
Where q = magnitude of charge
v = magnitude of the velocity vector = 4 x10^4 m/s
B = magnitude of the magnetic field vector
theta = the angle between the magnetic field and velocity vectors = 90°
m = mass of the charge = 0.195g
g = acceleration due to gravity =9.8m/s²
On substituting the respective values of all variables in the equation (1) above
B = 1.91T
The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.
Below is the step by step procedure to the solution.
