As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]
Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Answer:
N = 177843 sheets
Explanation:
We are given;
Mass;m = 0.0035 kg
Pressure; p = 101325 pa = 101325 N/m²
L = 0.279m
W = 0.216m
The weight of N sheets is N(mg)
Where;
m is the mass of one sheet
N is number of sheets
g is the acceleration due to gravity.
The pressure equals weight divided by the area on which the weight presses:
Thus,
p= F/A = Nmg/(L•W)
Therefore, making N the subject;
N = pLW/(mg)
N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)
N = 177843
Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>. That description is a little more direct.
It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".
Answer:
51.94 ft/s²
257.63 ft/s
Explanation:
t = Time taken = 4 s
u = Initial velocity = 34 mi/h
v = Final velocity
s = Displacement = 615 ft
a = Acceleration
Converting velocity to ft/s
Equation of motion
Acceleration is 51.94 ft/s²
Final velocity at this time is 257.63 ft/s
