Answer:
The answer is
. Let's learn why.
Explanation:
Newton's law of universal gravitation says;

Here G is a universal gravitational <u>constant</u> and is measured experimentally.
Sun's gravitational pull on mercury is:

Therefore 
Sun's gravitational pull on Earth is:

Therefore 
As a result;

Answer:
-4×-2y=14 (1)
-10×+7y=-25 (2)
multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2
-28×-14y=98
-20×+14y=-50
___________
-28×=48
×=48/-28
×=-12/7
now
-4×-2y=14
-4*-12/7-2y=14
48/7-2y=14
-2y=14-48/7
-2y=(98-48)/7
-2y=50/7
y=-50/14
y=-25/7
This question can be solved by using the equations of motion.
a) The initial speed of the arrow is was "9.81 m/s".
b) It took the arrow "1.13 s" to reach a height of 17.5 m.
a)
We will use the second equation of motion to find out the initial speed of the arrow.

where,
vi = initial speed = ?
h = height = 35 m
t = time interval = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,

<u>vi = 9.81 m/s</u>
b)
To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 17.5 m
vi = initial speed = 9.81 m/s
t = time = ?
Therefore,

solving this quadratic equation using the quadratic formula, we get:
t = -3.13 s (OR) t = 1.13 s
Since time can not have a negative value.
Therefore,
<u>t = 1.13 s</u>
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
I think it was Isaac Newton
Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight