Answer:
Moment of inertia of the system is 289.088 kg.m^2
Explanation:
Given:
Mass of the platform which is a uniform disk = 129 kg
Radius of the disk rotating about vertical axis = 1.61 m
Mass of the person standing on platform = 65.7 kg
Distance from the center of platform = 1.07 m
Mass of the dog on the platform = 27.3 kg
Distance from center of platform = 1.31 m
We have to calculate the moment of inertia.
Formula:
MOI of disk = ![\frac{MR^2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BMR%5E2%7D%7B2%7D)
Moment of inertia of the person and the dog will be mr^2.
Where m and r are different for both the bodies.
So,
Moment of inertia
of the system with respect to the axis yy.
⇒ ![I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g](https://tex.z-dn.net/?f=I_y_y%3DI_d_i_s_k%20%2B%20I_m_a_n%2BI_d_o_g)
⇒ ![I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2](https://tex.z-dn.net/?f=I_y_y%3D%5Cfrac%7BM_d_i_s_k%28R_d_i_s_k%29%5E2%7D%7B2%7D%20%2BM_m%28r_c%29%5E2%2BM_d_o_g%28R_c%29%5E2)
⇒ ![I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2](https://tex.z-dn.net/?f=I_y_y%3D%5Cfrac%7B129%281.61%29%5E2%7D%7B2%7D%20%2B65.7%281.07%29%5E2%2B27.2%281.31%29%5E2)
⇒
The moment of inertia of the system is 289.088 kg.m^2
Mercury and Venus are therefore closer to each other most of the time. But Earth is the planet closest to Venus. And that's why from here on Earth, Venus looks so big and luminous. Venus is the brightest thing in the night sky after the sun and the moon.
Answer: The force needed is 140.22 Newtons.
Explanation:
The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.
The force exerted on the pellet is
F = m a
In order to calculate the acceleration, given the displacement d,
![d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%5Cimplies%20a%3D%5Cfrac%7B2d%7D%7Bt%5E2%7D)
we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:
![\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}](https://tex.z-dn.net/?f=%5Coverline%7Bv%7D%3D%5Cfrac%7B1%7D%7B2%7D%28v_%7Bend%7D-v_%7Bstart%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%28155-0%29%5Cfrac%7Bm%7D%7Bs%7D%3D77.5%5Cfrac%7Bm%7D%7Bs%7D)
This value can be used to determine the time for the pellet through the barrel:
![t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7B%5Coverline%7Bv%7D%7D%3D%5Cfrac%7B0.6m%7D%7B77.5%5Cfrac%7Bm%7D%7Bs%7D%7D%5Capprox0.00774s)
Finally, we can use the above to calculate the force:
![F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%20m%5Cfrac%7B2d%7D%7Bt%5E2%7D%20%3D%200.007kg%5Ccdot%20%5Cfrac%7B2%5Ccdot%200.6%20m%7D%7B0.00774%5E2%20s%5E2%7D%5Capprox%20140.22N)
The answer is B.
This is because you add up all of the times (1.44s+1.70s+1.58s+1.76s) and you get 6.48 then you divide 6.48 by 4 to get the average of the times. Now you get the distance (200m) and because speed=distance/time you divide 200m/1.62s to get 123m/s. I hope this made sense :)
Answer:
A. velocity and time
Explanation:
A force can be define as an agent which has the capacity to change the state of an object. It can either increase the velocity of a body, change its direction of motion or cause a moving object to come to rest.
From Newton's second law of motion;
F = ma
where F is the force on the object, m is the mass of the object and a is the acceleration of the object. The unit of force is kgm/
or Newtons.
a = ![\frac{change in velocity}{change in time}](https://tex.z-dn.net/?f=%5Cfrac%7Bchange%20in%20velocity%7D%7Bchange%20in%20time%7D)
In the given question, apart from the mass of the object which is constant, the students should take the measurements of the velocity and time in each trial so as to calculate the required acceleration.