Answer:
7.62
Explanation:
because you have to divide 32/4.2
and can you do a friend request so i can accept it
Answer:
Explanation:
<u>2-D Projectile Motion</u>
In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are
The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:
Using the formulas for 
Simplifying
Dividing by 
Rearranging
Answer:
Fractional error = 0.17
Percent error = 17%
F = 112 ± 19 N
Explanation:
Plug in the values to find the force:
F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N
Find the fractional error:
ΔF/F = Δm/m + 2Δv/v + Δr/r
ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5
ΔF/F = 0.17
Multiply by 100% to find the percent error:
ΔF/F × 100% = 17%
Solve for the absolute error:
ΔF = 0.17 × 112 N = 19 N
Therefore, the force is:
F = 112 ± 19 N
Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°
