Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q = 
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( = 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
Hope this Helps!!!
Answer:
Explanation:
Given
Total time=27 min 43.6 s=1663.6 s
total distance=10 km
Initial distance 
time taken=25 min =1500 s
initial speed 
after 8.13 km mark steve started to accelerate
speed after 60 s
distance traveled in 60 sec
time taken in last part of journey
distance traveled in this time
and total distance
Answer:
Her error was that she did not subtract 12 from 8 correctly
Explanation:
Jackie did 8-12 instead of 12-8
The answer would be erin out of all of them thank me later :)
