Answer:
-34.5 W
Explanation:
rate of heat transfer = Stefan-Boltzmann constant x emissivity x surface area x temperature^4
![P = \alpha eA(T_{f} ^{4}- T_{i} ^{4}) = 5.67*10^{-8} *0.205*1.5(256^{4} - 284.4^{4} ) = -34.5 W](https://tex.z-dn.net/?f=P%20%3D%20%5Calpha%20eA%28T_%7Bf%7D%20%5E%7B4%7D-%20T_%7Bi%7D%20%5E%7B4%7D%29%20%3D%205.67%2A10%5E%7B-8%7D%20%2A0.205%2A1.5%28256%5E%7B4%7D%20-%20284.4%5E%7B4%7D%20%29%20%3D%20-34.5%20W)
When car is at the top of the hill its whole energy is stored in the form of gravitational potential energy
![U = mgh](https://tex.z-dn.net/?f=U%20%3D%20mgh)
so when height of the car becomes half then its potential energy is given as
![U_f = \frac{mgh}{2}](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7Bmgh%7D%7B2%7D)
so final potential energy when car falls down by half of the height will become half of the initial potential energy
So it is U = 50 MJ after falling down
Now by energy conservation we can say that final potential energy + final Kinetic energy must be equal to the initial potential energy of the car
So here at half of the height kinetic energy of car = 100 - 50 = 50 MJ
so we can say at this point magnitude of potential energy and kinetic energy will be same
<em>A. the same as the potential energy at that point.</em>
as we grow up we kinda experience some age staging so a teen got a lotta stuff going on they wanna be social active they be working on their popularity. so i guess thats it
Answer:
0.00321 Nm
Explanation:
= Final angular velocity = 3.5 rad/s
= Initial angular velocity = 0
= Angular acceleration
= Angle of rotation = 2 rev
Equation of rotational motion
![\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{3.5^2-0^2}{2\times 2\pi \times 1.8}\\\Rightarrow \alpha=0.54156\ rad/s^2](https://tex.z-dn.net/?f=%5Comega_f%5E2-%5Comega_i%5E2%3D2%5Calpha%20%5Ctheta%5C%5C%5CRightarrow%20%5Calpha%3D%5Cfrac%7B%5Comega_f%5E2-%5Comega_i%5E2%7D%7B2%5Ctheta%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cfrac%7B3.5%5E2-0%5E2%7D%7B2%5Ctimes%202%5Cpi%20%5Ctimes%201.8%7D%5C%5C%5CRightarrow%20%5Calpha%3D0.54156%5C%20rad%2Fs%5E2)
Moment of inertia is given by
![I=mr^2\\\Rightarrow I=0.25\times 0.154^2\\\Rightarrow I=0.005929\ kgm^2](https://tex.z-dn.net/?f=I%3Dmr%5E2%5C%5C%5CRightarrow%20I%3D0.25%5Ctimes%200.154%5E2%5C%5C%5CRightarrow%20I%3D0.005929%5C%20kgm%5E2)
Torque is given by
![\tau=I\alpha\\\Rightarrow \tau=0.005929\times 0.54156\\\Rightarrow \tau=0.00321\ Nm](https://tex.z-dn.net/?f=%5Ctau%3DI%5Calpha%5C%5C%5CRightarrow%20%5Ctau%3D0.005929%5Ctimes%200.54156%5C%5C%5CRightarrow%20%5Ctau%3D0.00321%5C%20Nm)
The torque required is 0.00321 Nm
As the plane falls the parabolic path remains directly below as the plane continues to fly over. This give more of an overview. When the package falls vertical acceleration happens as there is a vertical velocity as the package falls form high above. The downwards motion of gravity acts on the package if the approximated projectile motion ignoring air resistance.