To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as
Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is
At the same time since Newton's second law we know that:
Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to
Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:
But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:
Therefore there is a weight loss of 0.3N every kilometer.
If they both are moving with the same speed and direction
i.e. covering the same distance in the same time interval in the same direction
Net Force = (mass) x (acceleration) (Newton #2)
Net Force = (50 kg) x (6 m/s² down)
Net Force = (50 * 6) (kg-m/s² down)
<em>Net Force = 300 Newtons down</em>
Answer:
Correct answer: E total = 2,800 J
Explanation:
Given:
m = 4 kg the mass of the object
V = 20 m/s the speed (velocity) of the object
H = 50 m the height of the object above the surface
E total = ? J
The total energy of an object is equal to the sum of potential and kinetic energy
E total = Ep + Ek
Ep = m g H we take g = 10 m/s²
Ep = 4 · 10 · 50 = 2,000 J
Ek = m V² / 2
Ek = 4 · 20² / 2 = 2 · 400 = 800 J
E total = 2,000 + 800 = 2,800 J
E total = 2,800 J
God is with you!!!
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/
, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut + 
a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = a
= x 14 x 
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
