The ball rolled for 13.2 s
<h3>Further explanation</h3>
Speed is scalar and no direction
A bowling ball rolls 33 m, with average speed = 2.5 m/s
So elapsed time :
Answer:
The position of the particle is -2.34 m.
Explanation:
Hi there!
The equation of position of a particle moving in a straight line with constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the particle at a time t:
x0 = initial position.
v0 = initial velocity.
t = time
a = acceleration
We have the following information:
x0 = 0.270 m
v0 = 0.140 m/s
a = -0.320 m/s²
t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).
Then, we have all the needed data to calculate the position of the particle:
x = x0 + v0 · t + 1/2 · a · t²
x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²
x = -2.34 m
The position of the particle is -2.34 m.
Answer:
Lithification is the answer.
Answer:
a) Initial angular speed = 30 rad/s
b) Final angular speed = 70 rad/s
Explanation:
a) We have equation of motion s = ut + 0.5at²
Here s = 400 radians
t = 8 s
a = 5 rad/s²
Substituting
400 = u x 8 + 0.5 x 5 x 8²
u = 30 rad/s
Initial angular speed = 30 rad/s
b) We have equation of motion v = u + at
Here u = 30 rad/s
t = 8 s
a = 5 rad/s²
Substituting
v = 30 + 5 x 8 = 70 rad/s
Final angular speed = 70 rad/s
Answer
given,
Speed of car A = 95 Km/h
= 95 x 0.278 = 26.41 m/s
Speed of Car B = 121 Km/h
= 121 x 0.278 = 33.64 m/s
Distance between Car A and B at t=0 = 41 Km
a) Distance travel by car B
d = 26.41 t + 41000
speed of the car A = 33.64 m/s
distance = s x t
26.41 t + 41000 = 33.64 x t
7.23 t = 41000
t = 5670.82 s
time taken by Car B to cross Car A is equal to t = 5670.82 s
distance traveled by car A
D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km
b) distance travel by the car B in 30 s after overtaking car A
D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km
