Answer:
θ = 28°
Explanation:
For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.
X axis
Fₓ = m a
Nₓ = m a
Where the acceleration is centripetal
a = v² / r
The only force that we must decompose is normal, let's use trigonometry
sin θ = Nₓ / N
cos θ = 
/ N
Nₓ = N sin θ
= N cos θ
Let's replace
N sin θ = m v² / r
Y Axis
- W = 0
N cos θ = mg
Let's divide the two equations of Newton's second law
Sin θ / cos θ = v² / g r
tan θ = v² / g r
θ = tan⁻¹ (v² / g r)
We reduce the speed to the SI system
v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s
Let's calculate
θ = tan⁻¹ (16.94 2 / (9.8 55.1)
θ = tan⁻¹ (0.5317)
θ = 28°
Answer:
C. Third
Explanation:
it has completed just 2 1/2 orbits since its discovery
Gravitational force equals GMm/r^2, where G is constant, M and m are the masses, and r is distance.
For I, if both masses double, the formula becomes G2M2m/r^2, or 4GMm/r^2. Therefore, the gravitational force will quadruple or 4x.
For II, if the distance between the object doubles, the formula becomes GMm/(2r)^2 or GMm/4r^2. In this case, the gravitational force is 1/4x the initial force.
Answer:
No, the net force on the skydiver is zero
Explanation:
According to Newton's Second Law, the net force on an object is equal to the product between the mass of the object and its acceleration:
where
F is the net force
m is the mass of the object
a is the acceleration
In this problem, the acceleration of the skydiver is zero:
a = 0
This implies that also the net force on the skydiver is zero, according to the previous equation:
F = 0
So, the net force on the skydiver is zero. This occurs because the air resistance, which points upward, exactly balances the force of gravity on the skydiver, acting downwards.
Answer:
22.3 work is left to be done
2.4 work are done
