Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
Answer:
20m/s^2
Explanation:
Acceleration=Change in velocity/time taken for change
40-20/1
20m/s^2
Answer:
Explanation:
the center of mass formula
Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)
Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.
y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1
Ycm ( given ) = - .5
Putting the values of masses and positions
- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )
- .5 = 128 + 14 + m₃ x - 2.1 / ( 100+ m₃ )
- 50 - .5 m₃ = 142 - 2.1 m₃
1.6 m₃ = 192
m₃ = 120 kg .
B )
Total downward force is weight of total mass = 80 + 20 + 120
= 220 kg
weight = 220 x 9.8 = 2156 N .
component of weight perpendicular to rope
= 2156 cos 15 = 2082.53 N
This force will be equally distributed over each tree , so force on each tree = 2082.53 / 2 = 1041.26 N .
Answer:
One positive and one negative
D.) It is unlikely that a specific cause can be determined, but the treatment would likely be the same in either case.
