Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes


Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have


So total distance after which car will stop is


So car will not stop before the intersection as it is at distance 20 m
Answer:
False.
Explanation:
The statement shown in the question above is false and this can be confirmed by Newton's law on universal gravitation. According to Newton, the gravitational force exerted on any body is proportional to its weight, but the distance that the object travels when falling is disproportionate. In addition, if the force resulting from the weight of the object and its displacement has an angle of 0º, the weight force of that object will provide an increase in kinetic energy.
Answer:
newton - motion, gravity
kepler - orbital paths
brahe - the sun goes around the earth
Explanation:
im not sure about brahe but its the only one that makes sense
Answer:
0.54454
104.00902 N
Explanation:
m = Mass of wheel = 100 kg
r = Radius = 0.52 m
t = Time taken = 6 seconds
= Final angular velocity
= Initial angular velocity
= Angular acceleration
Mass of inertia is given by

Angular acceleration is given by

Equation of rotational motion

The coefficient of friction is 0.54454
At r = 0.25 m

The force needed to stop the wheel is 104.00902 N