Answer:
the ball didnt hit my face so
Explanation:
<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>
Explanation:
Let magnitude of the two forces be x and y.
Resultant at right angle R1= √15N) and at
60 degrees be R2= √18N.
Now, R1 = √(x² + y²) = √15,
R2= √(x² + y² +2xycos50) = √18.
So x² + y² = 15,
and x² + y² + 1.29xy = 18,
therefore 1.29xy = 3,
y = 3/1.29x.
y = 2.33/x
Now, x2 + (2.33/x)2 = 15,
x² + 5.45/x² = 15
multiply through by x²
x⁴ + 5.45 = 15x²
x⁴ - 15x2 + 5.45 = 0
Now find the roots of the equation, and later y. The two values of x will correspond to the
magnitudes of the two vectors.
Good luck
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
#SPJ1
Answer:
Average recoil force experienced by machine will be 200 N
Explanation:
We have give mass of each bullet m = 50 gram = 0.05 kg
There are 4 bullets
So mass of 4 bullets = 4×0.05 = 0.2 kg
Initial speed of the bullet u = 0 m/sec
And final speed of the bullet v = 1000 m/sec
So change in momentum 
Time is given per second so t = 1 sec
We know that force is equal to rate of change of momentum
So force will be equal to 
So average recoil force experienced by machine will be 200 N
