Answer:
accelerate / increase speed OR decelerate / decrease speed OR stop B1
change direction / move in a curve o.w.t.t.e
Explanation:
accelerate / increase speed OR decelerate / decrease speed OR stop B1
change direction / move in a curve o.w.t.t.e
Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash
Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.
Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her
Answer:
(a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Explanation:
Given that,
Radius = 5.9 m
(a). Angle 
We need to calculate the angle in radian
We need to calculate the path length
Using formula of path length
(b). Angle = 30 rad
We need to calculate the path length
(c). Angle = 30 rev
We need to calculate the angle in rad
We need to calculate the path length
Hence, (a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
An apple falling to the ground is not an example of centripetal acceleration.
