Answer:
The object distance is 20cm
Explanation:
Given
Magnification = -1.5
Image distance = 30 cm.
Required
Object Distance
We can calculate the object's distance using magnification formula;
M = -V/U
Where M = Magnification = -1.50
V = Image Distance = 30cm
U = Object Distance
Substitute the above parameters in the formula above.
-1.50 = -30/U
Multiply both sides by -1
-1.50 * -1 = -30/U * -1
1.50 = 30/U
Multiply both sides by U
1.50 * U = 30/U * U
1.50U = 30
Divide through by 1.50
1.50U/1.50 = 30/1.50
U = 30/1.50
U = 20cm
Recall that U represented the object distance.
Hence, the object distance is 20cm
The figure shows the arrangement of given data.
The displacement is considered along X direction and time along Y direction.
The slope of the graph is given by tan θ
So tan θ = Δx/Δt = Velocity between those two points considered for finding slope of graph.
So slope of position time is graph is equivalent to kinematic quantity Velocity.
Complete question:
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.
(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answer:
The net force on the person as the air bad deploys is -6750 N backwards
Explanation:
Given;
mass of the passenger, m = 60 kg
velocity of the car at impact, u = 15 m/s
final velocity of the car after impact, v = 0
distance moved as the front of the car crumples, s = 1 m
First, calculate the acceleration of the car at impact;
v² = u² + 2as
0² = 15² + (2 x 1)a
0 = 225 + 2a
2a = -225
a = -225 / 2
a = -112.5 m/s²
The net force on the person;
F = ma
F = 60 (-112.5)
F = -6750 N backwards
Therefore, the net force on the person as the air bad deploys is -6750 N backwards
B. Kinetic energy increase, gravitational potential energy decreases
