It is the the ecrins of light
Answer:
it appears to be farther away than it actually is, and therefore smaller then the object itself.
Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
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.