Answer and Explanation
Arranging the measured values in increasing order;
4.3s, 4.6s, 4.6s, 4.8s, 5.1s, 5.8s
The two outliers are obviously 4.3s and 5.8s; An outlier is a value in a statistical sample which does not fit a pattern that describes most other data point. Outliers make the average value complicated. So, it is usually better for data to be precise with data points spreading out around a small area.
So, the mean is the average of the four remaining data points after removing the outliers.
Mean = (4.6 + 4.6 + 4.8 + 5.1)/4
Mean = 4.775s
So, the value recorded should be 4.775s, 4.78s or 4.8s depending on the number of decimal places allowed.
QED!
Answer:
No, it is not conserved
Explanation:
Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.
The total kinetic energy before the collision is:

where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.
After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is

So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.
Answer:
2420 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 22.9 N
Angle (θ) = 35°
Distance (d) = 129 m
Workdone (Wd) =?
The work done can be obtained by using the following formula:
Wd = Fd × Cos θ
Wd = 22.9 × 129 × Cos 35
Wd = 22.9 × 129 × 0.8192
Wd ≈ 2420 J
Thus, the workdone is 2420 J.
Answer:
v =
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:

