A) Net horizontal force:
Fx= -35 + 25 + -10N
Net vertical force
Fy= 50 - 35 = 15N
Net force
F^2 = fX^2 +Fy^2
F= 18N
B.) Acceleration of system
a=18/5=3.6m/s^2
C). Equilibriant force must be equal to the net force above
F1= 18N
D.) Acceleration becomes half
a1= a/2 =1.8m/s^2
E.) Equilibrant force also doubles
F" = 2F' = 36N
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Answer:
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Answer:
The work done is 
Explanation:
Work is defined as the product of force and distance moved in the direction of application of force.
Given Data
distance S= 230m
force F= 140 N
Mass of car m= 1140 kg
Applying the formula for work done we have
The work done by pushing the car for a distance of 
is
Answer:
Therefore the horizontal range = 294897.96 m.
Explanation:
Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).
It can be expressed mathematically as
R = u²sin2∅/g............................. Equation 1
Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.
<em>Given: u = 1700 m/s, </em>∅ = 55°,
Constant: g = 9.8 m/s²
Substituting these values into equation 1
R = (1700²sin55)/9.8
R = 2890000/9.8
R = 294897.96 m.
Therefore the horizontal range = 294897.96 m.
Explanation :
Work is done when a force is applied to create a displacement on an object.
Thus, the work done depends on the two factors i.e.
(1) Applied force (F)
(2) Distance or displacement (d)
Mathematically, work done is 
It also depends on the angle between the force and the displacement.
For example,
A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:
or
So, 
Hence, the work done by him on the luggage is 294 Joules.
