Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
The answer ro this question is fear of failure
Answer:
The answer is 3.48 seconds
Explanation:
The kinematic equation
y= y0+V0*t+1/2*a*(t*t)
-50=0+(0)t+1/2(-9.8)*(t*t)
t=3.194 seconds
During ribbons ball,
x=x0+ Vt+1/2*a*(t*t)
x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)
x= 47.9157m
So, distance (D) = 100-47.9157= 52.084m
52.084m=0+15(t)+1/2*(0)(t*t)
t=52.084/15=3.472286= 3.48seconds
1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.
<span>
Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of
East. </span>
3. 1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.
hope this helps</span>
