Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
Answer:
one of the characteristics of a mammal is their several hollow bones another is their three chambered heart and the last is highly developed nervous system
Explanation:
the reason i picked those three is because not all mammals live their life on land and also mammals font have internal fertillization when they are done they take care of their babies and when they grow up they live their own life
Answer:
The Answer is D!
Explanation:
I checked it on Khan Academy.
Answer:
Mass of object is 0.5kg
Explanation:
Given the following data;
Force = 6N
Acceleration = 12m/s²
Mass =?
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
Making mass (m) the subject, we have;
Substituting into the equation;
Mass, m = 0.5kg.
Therefore, the mass of the object is 0.5kg
Answer:
0.54
Explanation:
Draw a free body diagram. There are 5 forces on the desk:
Weight force mg pulling down
Applied force 24 N pushing down
Normal force Fn pushing up
Applied force 130 N pushing right
Friction force Fnμ pushing left
Sum of the forces in the y direction:
∑F = ma
Fn − mg − 24 = 0
Fn = mg + 24
Fn = (22)(9.8) + 24
Fn = 240
Sum of the forces in the x direction:
∑F = ma
130 − Fnμ = 0
Fnμ = 130
μ = 130 / Fn
μ = 130 / 240
μ = 0.54
