<span>if the airplane moves at a speed of 25 m/s the mass is 855kg</span>
Answer:
Protons , neutrons and elections
<span>F* t = (m x v_final) - (m x v_initial)
200 x t = 50x8 - 50x0
t= 50 x 8 /200
t= 2s </span>
Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
(a) Let 
be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
where 
is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
so that
Solve for :
(b) The net force equation in part (a) leads us to the relation
so that is directly proportional to the square root of . As the radius increases, the maximum linear speed will also increase, so the cord is less likely to break if we keep up the same speed.
