initially, the car is traveling at 5.0 m/s.
so, we know acceleration for changing velocity is :
a = (v-v)/t ..........(i)
where v is the final velocity
v is the initial velocity
t is the time taken to change velocity
Now, as per the question :
initial velocity, v=5.0 m/s
final velocity, v =11 m/s
time taken, t = 3 s
putting the values in equation (i),
a = ( 11-5 )/3
a = 2 m/s²
Therefore, a, after 3 s, is <em>2 m/s².</em>
Answer:
a= 2.7 m/s^2
Explanation:
acceleration: a
speed: V
Vf = final speed
Vi= initial speed (initial = beginning)
100 km/hour --> m/s
divide the speed value by 3.6
100/3.6= 27.8 m/s
a= 2.7 m/s^2
The tension in each of the ropes is 625 N.
Draw a free body diagram for the bag of food as shown in the attached diagram. Since the bag hangs from the midpoint of the rope, the rope makes equal angles θ with the horizontal. The tensions <em>T</em> in both the ropes are also equal.
Resolve the tension T in the ropes into horizontal and vertical components T cosθ and T sinθ respectively, as shown in the figure. At equilibrium,
......(1)
Calculate the value of sinθ using the right angled triangles from the diagram.
Substitute the value of sinθ in equation (1) and simplify to obtain T.
Thus the tension in the rope is 625 N.
