Answer:
<h3>1.43m/s²</h3>
Explanation:
According to newtons second law.
F = mass * acceleration
If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg
Force applied = 1N
acceleration = Force/mass
Substitute the values and get acceleration
acceleration = 1/0.7
acceleration = 1.43m/s²
Hence the magnitude of the acceleration of the robot is 1.43m/s²
Answer:
A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.
So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.
So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle
As we know that arc length is given as

length of chord is given as

so here


so we have

<span>b.) Add heat to increase the molecular motion of the atoms</span>
Answer:
The velocity of the motorboat after 6s is 24 m/s.
Explanation:
Given;
acceleration of the motorboat, a = 4.0 m/s²
initial velocity of the motorboat, u = 0
time of motion of the motorboat = 6s
Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;
v = u + at
v = 0 + (4 x 6)
v = 24 m/s
Therefore, the velocity of the motorboat after 6s is 24 m/s.
Answer:
W = 2352 J
Explanation:
Given that:
- mass of the bucket, M = 10 kg
- velocity of pulling the bucket, v = 3

- height of the platform, h = 30 m
- rate of loss of water-mass, m =

Here, according to the given situation the bucket moves at the rate,

The mass varies with the time as,

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance
∆x = 3∆t meters
So, during this interval change in work done,
∆W = m.g∆x
<u>For work calculation:</u>
![W=\int_{0}^{10} [(10-0.4t).g\times 3] dt](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B10%7D%20%5B%2810-0.4t%29.g%5Ctimes%203%5D%20dt)
![W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}](https://tex.z-dn.net/?f=W%3D%203%5Ctimes%209.8%5Ctimes%20%5B10t-%5Cfrac%7B0.4t%5E%7B2%7D%7D%7B2%7D%5D%5E%7B10%7D_%7B0%7D)
