Answer:2800000j
Explanation:
For us to know the kinetic energy of the vehicle,
Where m is the mass
And v is the velocity
Then, K.E=1/2mv^2
While, K.E=1/2×3500×40^2
Therefore, our answer will now be
K.E=2800000j
Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
The answer to this question would be: vitamin D
The UV is needed by the skin to make previtamin D3. Previtamin D3 or cholecalciferol made from the skin will be changed in the liver into 25- hydroxyvitamin D3 and then sent to the kidney to be changed into 1,25- dihydroxyvitamin D3. The last change in the kidney will active the vitamin D. Vitamin D has a role in the calcium absorption which was will strengthen the bone tissue.
Answer:
.
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is
(same as
,) meaning that
of energy is supplied to this lamp every second.
The question states that
out of that
of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the
of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive
of energy input and would outputs
of useful work. The efficiency of this lamp would be:
.