Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer:
2.5m/s2
Explanation:
The following were obtained from the question:
F = 600N
M = 240 kg
a =?
Recall: F = Ma
a = F/M
a = 600/240
a = 2.5m/s2
Therefore, the acceleration of the motorcycle is 2.5m/s2
Answer:
38.87 m/s
Explanation:
Given that the ball is dropped from a height = 77 m
u = 0 m/s
s = 77 m
a = g = 9.81 m/s²
Applying the expression as:

Applying values as:

<u>The speed with which the ball hit the ground = 38.87 m/s</u>
Answer:
20.7 s
Explanation:
The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:
v = a*t + v₀
Solve for t:
t = (v - v₀)/a