Answer:
The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².
How many times stronger than gravity is this force? 2.79 x 10⁴ g.
Explanation:
given information:
s = 220 m
final speed, vf = 10.97 km/s = 10970 m/s
g = 9.8 m/s²
he unrealistically large acceleration experienced by the space travelers during their launch
vf² = v₀²+2as, v₀ = 0
vf² = 2as
a =vf²/2s
= (10970)²/(2x220)
= 2.7 x 10⁵ m/s²
Compare your answer with the free-fall acceleration
a/g = 2.7 x 10⁵/9.8
a/g = 2.79 x 10⁴
a = 2.79 x 10⁴ g
Answer:
Electric current is electric charge in motion. It can take the form of a sudden discharge of static electricity, such as a lightning bolt or a spark between your finger and a ground light switch plate. ... Most electric charge is carried by the electrons and protons within an atom.
Explanation:
because it is
Answer:
Please see below as the answer is self-explanatory.
Explanation:
- We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
- Both movements are independent each other, due to they are perpendicular.
- In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
- Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

- Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

- In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
- Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

- In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

- Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

- Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
Answer:
15 to 30 so anywhere in between there.
<span>as i recall, gravity is relative to the square of the distance.
so if the distance is tripled, then the gravitational attraction would be reduced by 3^2 or 1/9.
so F1 = F0/9
if the satellite is 2R from the center, and is moved to 4R (doubled would be 3R, tripled is 4R) then the distance is twice, and gravity would be 2^2 or 1/4.
</span>