Given that,
Initial velocity , Vi = 0
Final velocity , Vf = 40 m/s
Acceleration due to gravity , a = 9.81 m/s²
Distance can be calculated as,
2as = Vf² - Vi²
2 * 9.81 *s = 40² - 0²
s = 81.55 m
For half height, that is, s = 40.77m
Vf= ??
2as = Vf² - Vi²
2 * 9.81 * 40.77 = Vf² - 0²
Vf² = 800
Vf = 28.28 m/s
Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.
Answer:
the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.
Explanation:
Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:
Then, if we assign the subindex "1" to the quantities that define the magnetic field (
) inside solenoid 1, we have:
![B_1=\mu_0\, \frac{N_1}{L_1} I_1](https://tex.z-dn.net/?f=B_1%3D%5Cmu_0%5C%2C%20%5Cfrac%7BN_1%7D%7BL_1%7D%20I_1)
notice that there is no dependence on the diameter of the solenoid for this formula.
Now, if we write a similar formula for solenoid 2, given that it has :
1) half the length of solenoid 1 . Then ![L_2=L_1/2](https://tex.z-dn.net/?f=L_2%3DL_1%2F2)
2) twice as many turns as solenoid 1. Then ![N_2=2\,N_1](https://tex.z-dn.net/?f=N_2%3D2%5C%2CN_1)
3) three times the current of solenoid 1. Then ![I_2=3\,I_1](https://tex.z-dn.net/?f=I_2%3D3%5C%2CI_1)
we obtain:
![B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1](https://tex.z-dn.net/?f=B_2%3D%5Cmu_0%5C%2C%20%5Cfrac%7BN_2%7D%7BL_2%7D%20I_2%5C%5CB_2%3D%5Cmu_0%5C%2C%20%5Cfrac%7B2%5C%2CN_1%7D%7BL_1%2F2%7D%203%5C%2CI_1%5C%5CB_2%3D%5Cmu_0%5C%2C%2012%5C%2C%5Cfrac%7BN_1%7D%7BL_1%7D%20I_1%5C%5CB_2%3D12%5C%2CB_1)
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
#SPJ1
Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.
Heat stroke had occurred when your body can no longer regulate its temperature. hope it helps :)