Answer:
The motorcycle travelled 69.73 m during these 3.1 s.
Explanation:
In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:
a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2
The distance is given by the following formula:
S = (V_final^2 - V_initial^2)/(2*a)
S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m
The motorcycle travelled 69.73 m during these 3.1 s.
Answer:
The acceleration of the collar is 10 m/s²
Explanation:
Given;
mass of the collar, m = 1 kg
applied force on the bar, F = 10 N
The acceleration of the collar can be calculated by applying Newton's second law of motion;
F = ma
where;
F is the applied force
m is mass of the object
a is the acceleration
a = F / m
a = 10 / 1
a = 10 m/s²
Therefore, the acceleration of the collar is 10 m/s²
Answer:
0.687 m/s
Explanation:
Initial energy = final energy
1/2 mu² = mgh + 1/2 mv²
1/2 u² = gh + 1/2 v²
Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:
1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²
v = 0.687 m/s
Answer:
(a)11.24 m/s
(b)7.44 m/s
(c)409 N
(d)
(e) 0
Explanation:
The period for 1 circle of the merry go around is 9.5s. It means the angular speed is:
(a)The speed is
(b) Centripetal acceleration:
(c) Magnitude of the force that keeps you go around at this acceleration
(d) let the coefficient of friction by . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man