a) Pull project 2 (the left) shows constant velocity. For each Second elapsed the distance traveled increases by the same amount, consistent with Amy traveling a constant distance/sec. The derivative of an identical equation would be a flat line, and the derivative of distance traveled is velocity. A flat line in velocity is a constant velocity.
b)Pull project 1 (the right) shows constant acceleration. As time passes, the distance traveled in each elapsed time increases. The curve looks like one side of a quadratic equation, its derivative would be a linear equation. The derivative of distance traveled over time is velocity, a constant increase in speed is a constant acceleration.
c)Pull Project 2 shows that being pulled with a constant force results in a constant velocity, since the graph of data indicates a constant velocity.
d)Pull Project 1 shows that being pulled with a constant force results in a constant acceleration, since the graph of the data indicates a constant acceleration.
e)Pull Project 1 would be Amy on the near frictionless cart. Force=Mass x Acceleration, since Amy's mass doesn't change, the constant force has to effect a constant acceleration.
f)Pull Project 2, where the velocity remains constant indicates that there is a force opposing the force pulling Amy resulting in no net acceleration once she starts moving. Friction from Amy sliding along the floor could be that counter force. Pull Project 1, where the acceleration is constant indicates no counter force, i.e. no friction.
The second law of Newton that relates the force to the acceleration and mass of an object is:
`F = m*a`
Thus, the inverse of the slope of the graph should be equal to the mass of the object. For object 2 the inverse of the slope is
`m_2 =1/s_2 = (x_2/y_2) =5/(5a1) =1/(a1)`
`95 =1/(a1)` , thus `a1 =1/95 =0.0105 m/s^2`
For objects 1 and 3 the inverses of the slopes are
`m_1=1/s_1 =(x_1/y_1) =2/(5*a1) =2/(5*0.0105) =38.095 =38.1 kg`
`m_3 =1/s_3 =(x_3/y_3) = 5/(2a1) =5/(2*0.0105) =238.095 kg =238.1 kg`
The masses of objects 1 and 3 are 38.1 kg and respectively 238.1 kg.
You need to evaluate the magnitude of centripetal force that keeps Mars in its orbit, such that:
`F_c = m*a_c`
`m` represents the mass of Mars
`a_c` represents the centripetal acceleration
Since the mass of Mars is provided by the problem, you need to evaluate the centripetal acceleration, such that:
`a_c = v^2/r`
`r` represents theradius of circular orbit
`v` represents the mean orbital velocity
Since the problem provides the values of `v` and `r` , you need to replace its values in equation, such that:
`a_c = 24.13^2*10^6/2.28*1011`
`a_c = 0.252589*10^6 m*s^(-2) => a_c = 252589 m*s^(-2)`
Replacing the value of centripetal acceleration in equation that helps you to evaluate the centripetal force, yields:
`F_c = m*a_c => F_c = 6.42*1023*252589 => F_c = 1658.9186*10^(-6) N`
Hence, evaluating the magnitude of centripetal force that keeps Mars in its orbit, yields `F_c = 1658.9186*10^(-6) N` .