Answer:
a = 2.22 [m/s^2]
Explanation:
First we have to convert from kilometers per hour to meters per second
![40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]](https://tex.z-dn.net/?f=40%20%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A%5B%5Cfrac%7B1h%7D%7B3600s%7D%5D%2A%5B%5Cfrac%7B1000m%7D%7B1km%7D%5D%20%3D%2011.11%20%5Bm%2Fs%5D)
We have to use the following kinematics equation:
where:
Vf = final velocity = 11.11 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 5 [s]
The initial speed is taken as zero, as the car starts from zero.
11.11 = 0 + (a*5)
a = 2.22 [m/s^2]
Answer:
a) Total distance travelled = 1.9km + 8.7km + 3.0km = 13.6km
b) Total displacement = -1.9km + 8.7km-3.0km=+3.8km
Total displacement = 3.8km due west
Explanation:
a) Total distance travelled is a scalar quantity it is not affected by the direction it's only concerned with the magnitude.
Total distance travelled = 1.9km + 8.7km + 3.0km = 13.6km
b) Total displacement is a vector quantity it takes into consideration both magnitude and direction.
Taking west as positive and east as negative.
Total displacement = -1.9km + 8.7km-3.0km = +3.8km
Total displacement = 3.8km due west
Sorry I don’t know but good luck and hope you find it
Answer:
v = 2.348 m/s
Explanation:
Let:
h be the height of the child above the swing's low point,
v be the speed at the low point.
When the ropes make angle 3
25 deg. with the vertical:
h = 3.00(1 - cos(25)).
Equating potential energy at the high point to kinetic energy at the low point:
29.0 * 9.81 * 3.00(1 - cos(25)) = 29.0v^2 / 2
79.96 =29.0v^2 / 2
v = 2.348 m/s
