Answer:
Here is the solution hope it helps:)
Answer:
Acceleration=24.9ft^2/s^2
Angular acceleration=1.47rads/s
Explanation:
Note before the ladder is inclined at 30° to the horizontal with a length of 16ft
Hence angular velocity = 6/8=0.75rad/s
acceleration Ab=Aa +(Ab/a)+(Ab/a)t
4+0.75^2*16+a*16
0=0.75^2*16cos30°-a*16sin30°---1
Ab=0+0.75^2sin30°+a*16cos30°----2
Solving equation 1
(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s
Also from equation 2
Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2
<span>its kinetic energy is 7350kJ
</span>
Kinetic energy is given as =

Now, m = 12 gms = 0.012 kg
And, velocity = 35 kilometers/second = 35000 m/sec
Kinetic energy is given as =
![\frac{1}{2} 0.012 kg * 35000*35000 m/[tex] s^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%200.012%20kg%20%2A%2035000%2A35000%20m%2F%5Btex%5D%20s%5E%7B2%7D%20)
= 6

×1225 ×

m/
= 7350 kJ
Answer:
a. 900 J
b. 0.383
Explanation:
According to the question, the given data is as follows
Horizontal force = 150 N
Packing crate = 40.0 kg
Distance = 6.00 m
Based on the above information
a. The work done by the 150-N force is


= 900 J
b. Now the coefficient of kinetic friction between the crate and surface is


= .383
We simply applied the above formulas so that each one part could calculate