Probably to the right because the force on the left is greater and will therefore overpower the 2N force and push to the right
Vi = vf + ( a.t)
0 m/s (rest) = 80 m/s + (a.20s.-1)
a.20.-1= -80, we check if this is true; = 0 m/s= 80 + -80 = 0 ITS TRUE.
so a.20.-1= -80, a= 80/20 , Answers a= 4 m/s
Answer:
e) 31.6 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion




Time taken by the cars to meet 31.6 seconds.
Answer:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I =
)^{2}[/tex]
I =
)^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I
= 0.5 x 0.07875 x
= 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
Answer:
40 s
Explanation:
After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start
Let t be the number of seconds after the second skater starts will the second skater overtake the first skater
The distance traveled by the first skater after t seconds is

Similarly the distance traveled by the 2nd skater after t seconds is

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

We can substitute 


