Fnet =ma
1560)(1.3102)
the answer is b
Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of 
Coordinate of position of 
Time = 2.00 s
We need to calculate the acceleration

For x coordinates

On differentiate w.r.to t

On differentiate again w.r.to t

The acceleration in x axis at 2 sec

For y coordinates

On differentiate w.r.to t

On differentiate again w.r.to t

The acceleration in y axis at 2 sec

The acceleration is

We need to calculate the net force



The magnitude of the force


Hence, The net force acting on this object is 180.89 N.
Velocity is define as the time rate taken for a change in acceleration
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
Answer:
Your answer is the following...
(b)"Many valence electrons are shared between the atoms"
Explanation: