It’s North of the equator
Answer:
mass*velocity=1.5*10^4 * 15
= 22.5*10^4
That's 1/2 of what it is on the surface.
The distance between the center of the Earth and any object
on the surface is 1 Earth radius ... about 3960 miles.
Gravitational force is inversely proportional to the square of
the distance between the centers of the objects, so in order
to reduce the acceleration of gravity by 1/2, you increase the
distance by √2 .
(3960 miles) x (√2) = 5,600 miles from the center
= 1,640 miles above the surface.
Answer:
<h2>a) 496N</h2><h2>b) 50.56kg</h2><h2>c) 1.90m/s²</h2>
Explanation:
According to newton's secomd law, ∑F = ma
∑F is the summation of the force acting on the body
m is the mass of the body
a is the acceleration
Given the normal force when the elevator starts N1 = 592N
Normal force after the elevator stopped N2 = 400N
When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)
When moving up;
N1 - Fg = ma
N1 = ma + Fg ...(1)
Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;
N2 - Fg = -ma
N2 = -ma+Fg ...(2)
Adding equation 1 and 2 we will have;
N1+N2 = 2Fg
592N + 400N = 2Fg
992N 2Fg
Fg = 992/2
Fg = 496N
The weight of the person is 496N
<em>\b) To get the person mass, we will use the relationship Fg = mg</em>
g = 9.81m/s
496 = 9.81m
mass m = 496/9.81
mass = 50.56kg
c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;
N1-N2 = 2ma
592-400 = 2(50.56)a
192 = 101.12a
a = 192/101.12
a = 1.90m/s²
Answer:


Explanation:
<u>Given:</u>
Mass = m = 200,000 kg
Vertical Distance = h = 120 m
Speed = v = 14 m/s
Acceleration due to gravity = g = 10 m/s²
<u>Required:</u>
1) Gravitational Potential Energy = P.E = ?
2) Kinetic Energy = K.E. = ?
<u>Formula:</u>
1) P.E. = mgh
2) K.E. = 
<u>Solution:</u>
1) P.E. = (200,000)(10)(120)
P.E. = 240,000,000 Joules
P.E. = 240 Mega Joules
P.E. = 240 MJ
2) K.E. = 1/2 (200000)(14)^2
K.E. = (100000)(196)
K.E. = 19,600,000 Joules
K.E. = 19.6 MJ
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>