Answer:
The answer to the question is
The object would fall 57.625 m in the first 5 seconds
Explanation:
To solve the question, we note that
the height of fall = 490 ft = 149.352 m
Time to touch the ground = 7 seconds
We are required to find out how far the object falls in the first 5 seconds
We apply the relation
S = u·t + 0.5×g·t ² = We then have
149.352 = U×7+0.5*9.81*49 From where u = -13 m/s
Therefore to find how far it falls in the first 5 seconds, we have
-13*5 + 0.5*9.81*25 = 57.625 m
Answer:
Explanation:
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Answer:
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object
Explanation:
it's newton's second law
Magnitude of normal force acting on the block is 7 N
Explanation:
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
= 30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N
Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg
