To solve this problem we will use Newton's second law in order to obtain the weight of a person. The second law tells us that
F = ma
Where,
m = mass
a = Acceleration
In this particular case, the acceleration is equal to that exerted by the earth through gravitational acceleration, so if the person's weight is 75Kg and the gravity is 
, the weight of the body will be,
When the elevator is at rest this reads 735.75N and 75Kg (The same mass of the person)
An educated guess to something you should test over and over
Answer: North
Explanation: I believe the friction will go the opposite way of the object being pushed.
Answer:
a) r = 4.22 10⁷ m, b) v = 3.07 10³ m / s and c) a = 0.224 m / s²
Explanation:
a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force
F = m a
a = v² / r
F = G m M / r²
G m M / r² = m v² / r
G M / r = v²
The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships
v = d / t
As the orbit is circular the distance is the length of the circle in 24 h time
d = 2π r
t = 24 h (3600 s / 1 h) = 86400 s
Let's replace
G M / r = (2π r / t)²
G M = 4 π² r³ / t²
r = ∛(G M t² / (4π²)
r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)
r = 4.22 10⁷ m
b) the speed module is
v = √G M / r
v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷
v = 3.07 10³ m / s
c) the acceleration is
a = G M / r²
a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²
a = 0.224 m / s²
Explanation:
Given:
v₀ = 0 m/s
v = 49 m/s
a = 9.8 m/s²
Find: t
v = at + v₀
49 m/s = (9.8 m/s²) t + 0 m/s
t = 5 s
