Answer:
(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).
(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .
Explanation:
m= 3kg
a= 2 i + 5 j = 5 .38 < 68.19 º
F= m * a
F= 3* ( 5.38 < 68.19º )
F= 16.4 N < 68.19º
Fx= F * cos(68.19º)
Fx= 5.99
Fy= F* sin(68.19º)
Fy= 14.98
Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s
We have to find final velocity.
The equation we use is
Final velocity=initial velocity+acceleration x time
Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s
We would round this to:
Vf (final velocity)=2.7m/s
Answer:
(A) She needs to move the decimal point by 3 places
Answer:
Explanation:
From the question we are told that:
Height 
Time 
Generally the Newton's equation for Initial velocity upward is mathematically given by
Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28
Generally the Newton's equation for time to reach initial velocity is mathematically given by
