Answer:
μ = 0.109
Explanation:
Draw a free body diagram of the crate. There are four forces:
Weight force mg pulling down.
Normal force N pushing up.
Applied force P pulling at θ above the horizontal.
Friction force Nμ pushing to the left.
Sum of the forces in the y direction:
∑F = ma
N + P sin θ − mg = 0
N = mg − P sin θ
Sum of the forces in the x direction:
∑F = ma
P cos θ − Nμ = ma
P cos θ − ma = Nμ
μ = (P cos θ − ma) / N
μ = (P cos θ − ma) / (mg − P sin θ)
Given:
P = 585 N
θ = 28.0°
m = 125 kg
a = 3.30 m/s²
μ = (585 cos 28.0° − 125 kg × 3.30 m/s²) / (125 kg × 9.8 m/s² − 585 sin 28.0°)
μ = 0.109
Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Answer:
Turn the heater on
Explanation:
There are two main forces involved in a balloon flight
The downward force is the total weight of the balloon: the air it contains, the gas bag, the basket, the passengers, etc.
The upward force is the weight of the of the air the balloon displaces.
During level flight
,
buoyant force = weight of displaced air - total weight of balloon
If you increase the temperature of the air in the bag, the air molecules spread out and leave through the bottom of the bag.
The balloon still has the same volume, so the weight of displaced outside air stays the same.
However, the balloon has lost some hot inside air, so its total weight decreases.
The upward force is greater than the downward force, so the balloon rises.
Horizontal component = (10N) · sin (20°) = 3.42... N (rounded)
Vertical component = (10N) · cos (20°) = 9.39... N (rounded)
