A horizontal force of 0.9N act on a body of mass 0.1kg there is resistance of 0.15 Newton opposing the first Force what accelera
1 answer:
You might be interested in
Answer:
The resultant tension of the two ropes is approximately 42.4 N
The length of the line representing the resultant tension is approximately 8.48 cm
Please find included with the answer the scale drawing created with Microsoft Word
Explanation:
The given parameters are;
The tension in rope P, = 30 N
The tension in rope Q, = 30 N
The angle the rope, 'P', makes with the horizontal = 45°
The angle the rope, 'Q', makes with the horizontal = 45°
The scale factor of the scale diagram, S.F. = 5.0 N/cm
By the resolution of forces at equilibrium, we have;
The sum of the vertical forces, =
+
+ W = 0
∴ W = -( +
)
W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712
The weight of the heavy ball, W ≈ 42.4 N acting downwards
The sum of the horizontal forces, =
+
= 0
The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm
The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile: