Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
Answer:
293.7 degrees
Explanation:
A = - 8 sin (37) i + 8 cos (37) j
A + B = -12 j
B = a i+ b j , where and a and b are constants to be found
A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
Comparing coefficients of i and j:
a = 8 sin (37) = 4.81452 m
b = -12 - 8cos(37) = -18.38908
Hence,
B = 4.81452 i - 18.38908 j ..... 4 th quadrant
Hence,
cos ( Q ) = 4.81452 / 12
Q = 66.346 degrees
360 - Q = 293.65 degrees from + x-axis in CCW direction
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.
Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
<span>x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v=<span>
<span>-0.13261622 m/s
</span></span>the object moving at 0.13 m/s <span>at time t=0.815 s</span>
C) electrical energy is transformed into heat energy
