<u>Answer:</u>
Total displacement traveled = 298
<u>Explanation:</u>
According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).
So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.
Then after a little it can travel up another cm to reach the top.
Therefore, the total displacement traveled = 297 + 1 = 298 cm
Answer:
The velocity of the ball after 5 seconds will be 49 m/s
Explanation:
<em>v = final velocity</em>
<em>u = initial velocity</em>
<em>g = acceleration due to gravity</em>
<em>t = time</em>
Initial velocity of the ball = 0 (As the ball is dropped from rest )
Acceleration due to gravity = 9.8 m/s
Time taken = 5 sec
As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question
Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v
Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.
Velocity of the ball after 5 seconds
Distance covered by the ball in 5 sec
Answer:
The acceleration of the crate is 
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
Answer:
B) 2.7W
Explanation:
Converting Cal to Joule
1 cal = 4.186J
14 kcal = 14 x 1000 x 4.186
= 58604 J
Converting hour to seconds
6 hours = 6 x 60 x 60 seconds
= 21600 seconds
Power is the time rate of doing work.
Power = Work/Time
P = (58604) / (21600)
P = 2.7W
The work done by 
along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,
I assume the path itself is a line segment, which can be parameterized by
with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is
![\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cleft%286x%28t%29%5E3%5C%2C%5Cvec%5Cimath-4y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%29%5Ccdot%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Bx%28t%29%5C%2C%5Cvec%5Cimath%20%2B%20y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%5D%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%28288%283t-1%29%5E3-8%282t%2B5%29%29%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B312%7D)
