The particle has a constant horizontal velocity, and a vertical force won't affect the horizontal speed, so it should be fairly easy to find the last part, "the time taken for a 10m horizontal displacement," using a kinematic equation.
X = x + vt + (1/2)at²
10 = 0 + (1.6)t + (1/2)(0)t²
10/1.6 = t
t = 6.25s
So now we have to find the vertical displacement over 6.25 seconds on a particle of a 2.5kg mass with a force of 8N.
Start with Newton's second law.
F = ma
8 = (2.5)a
a = 3.2m/s²
Now, use kinematics again.
Y = y + vt + (1/2)at²
Y = 0 + (0)(6.25) + (1/2)(3.2)(6.25)²
Y = <u>62.5m</u>
Answer: M = 6.13 × 10^18 kg
Explanation:
g = GM/r2,
Where
The mass M of the asteroid = ?
The radius r = 110000 m
g = 0.0338 m/s^2
G is the gravitational constant.
SI units its value is approximately 6.674×10^−11m3⋅kg−1⋅s−2
Using the formula
g = GM/r2
Cross multiply
GM = gr^2
6.674×10^-11M = 0.0338 × 110000^2
M = 408×10^6/6.674×10^-11
M = 6.13 × 10^18 kg
Answer:
1: body of significant size
Explanation:
Hopefully this helps!