Answer:
I would measure the distance between points A and B
Answer:
a) u = 6 m/s
b) a = 4 m/s²
c) d(3) = 16 m
Explanation:
equation for the first second
distance will be the average velocity times the time of travel
8 = ½(u + (u + at))t t is one second, so reduces to
8 = u + ½a
velocity at the end of the first second is
v = u + at = u + a
position equation for the second period is
12 = ½((u + a) + (u + a + at))t t is one second so reduces to
12 = u + 3a/2
subtracting the first position equation from the second
12 - 8 = u + 3a/2 - (u + ½a)
a = 4 m/s²
8 = u + ½4
u = 6 m/s
in the third second
d = 6(3) + ½(4)(3²) - 8 - 12
d = 16 m
Answer:
Magnitude of Force : 0.8 N
Explanation:
We know that the velocity of this particle changes from 15ms⁻¹, or in other words 15m / s, to the respective velocity 25m / s over the course of 2.5 seconds. Given this information we can determine the acceleration of the particle,
a = v₁ - v₂ / t = 25 - 15 / 2.5 = 10 / 2.5 = 4m / s²
Knowing the acceleration we can calculate the magnitude of the force using the formula f = ma - Newton's second law of motion,
f = m a = 200g 4m / s² = 800 g m / s²
Remember however that Newtons are in the standard units kg m / s². Therefore we have to convert 800 g to kg to receive our solution,
800 g = 800 / 1000 kg = 0.8 kg,
Magnitude of Force = 0.8 N ; Solution = Option B
Answer:
Explanation:
Spilling the force of 60N on the rope at an angle θ in x-y direction:
fx = Fcosθ = 60cosθ
fy = Fsinθ = 60sinθ
cosθ and sinθ both vary from 0 to 1 between 0° to 90° but they go in opposite direction:
cos0° = 1 and cos90° = 0
sin0° = 0 and sin90° = 1
sin45° = cos45° = 0.7071
When θ increases from 0° to 90°,
θ=0°, fx = 60cos0° = 60N, fy = 60sin0° = 0N
θ=90°, fx = 60cos90° = 0N, fy = 60sin90° = 60N
θ=45°, fx = 60cos45° = fy = 60sin45° = 42.43N