Answer:
Explanation:
The package had the same velocity as the plane when it was dropped. Newton's 1st Law says that "an object in motion tends to stay in motion, at the same velocity, in a straight line unless acted on by an outside force".
There only outside force acting on the package was its weight -- that force is straight down. The horizontal velocity that the plane gave the package continued (as Newton said it would), so as it fell, horizontally it kept pace with the plane.
<span>b) The force with a distance of 150 km is 889 N
c) The force with a distance of 50 km is 8000 N
This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question.
Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as
F = (G M1 M2)/r^2
where
F = Force
G = gravitational constant
M1 = Mass 1
M2 = Mass 2
r = distance between center of masses for the two masses.
So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889.
Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits.
If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
Answer:
1845.26 ?
Explanation:
18.46 × 99.96= 1845.2616 = 1845.26
im not entirely sure though
Answer:
The acceleration due to gravity at Pluto is 0.0597 m/s^2.
Explanation:
Length, L = 1 m
10 oscillations in 257 seconds
Time period, T = 257/10 = 25.7 s
Let the acceleration due to gravity is g.
Use the formula of time period of simple pendulum
