Was the ball aded eastward at a speed of 156 m/s. A 20.0 m/s wind is blowing tim

Question

Was the ball aded eastward at a speed of 156 m/s. A 20.0 m/s wind is blowing time as the plane is flying. What is the velocity of the plane southward at the same relative to the ground (Magnitude and Direction)? 5. A 6 0.0-kg person rid acceleration 2.00 m/s2 upward. What is the reading on the scale? 6. A ball is thrown at a 40.0 angle above the horizontal across level ground. Itis released from a height of 2.00 m above the ground with a speed of 18.0 m/s. What is the maximum height above the ground that the ball reaches? 7. Two forces act on a 4.00-kg object in a manner that the object has an acceleration 3.00 m/s2 in a direction 20.0° north of east. The first force is 15.00 N in a direction 10.0° west of north. What is the second force (Magnitude and Direction)? 8. Vectors A and B are shown below. What is 0044.0 ? .2

Explanation / Answer

#5)

along vertical Fnet = N - W

from newtons second law Fnet = m*a

N - W = m*a

N = W + m*a

N = m*g + m*a

N = m*(g+a)

N = 60*(9.8+2) = 708 N

reading m' = N/g = 708/9.8 = 72.24 kg

===============

7)

resultant force R = 4*3 = 12 N

components of R

Rx = R*cos20 = 12*cos20 = 11.3 N

Ry = R*sin20 = 12*sin20 = 4.1 N

F1x = -15*sin10 = -2.6 N

F1y = 15*cos10 = 14.8 N

F2x = ?

F2y = ?

Rx = (F1x+F2x)

F2x = Rx - F1x = 11.3 + 2.6 = 13.9 N

Ry = F1y + F2y

F2y = Ry - F1y = 4.1 - 14.8 = -10.7 N

magnitude of second force F2 = sqrt(F2x^2+F2y^2)

F2 = sqrt(13.9^2+10.7^2) = 17.5 N

direction = tan^-1(F2y/F2x) = 37.5 degrees south of west <<<<-----------ANSWER

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