The result of the Sun's light being refracted when it is
high in the sky around noon is blue, indigo and violet are strongly bent that
mostly skim through the upper atmosphere making the sky appear blue. The
refraction of the light during noon makes the sky blue.
You can use the Pythagorean theorem to calculate the hypotenuse, since the resultant is the hypotenuse of the triangle.
Answer:
Coefficient of dynamic friction= md= 0.09931
Explanation:
To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:
F= 90N × cos(30°) = 77.9423N
This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:
F= md × m × g= 77.9423N
m= mass of the block
md= coefficient of dynamic friction
g= gravity acceleration
F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)
md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931
Answer:
5 ) The mass, 6) with lubrication and using surfaces that are not rough
Explanation:
5) If two bodies are held regardless of their densities and can be combined by some chemical or physical process, the only physical property to be modified will be the mass of the resulting body.
8)
Friction depends on the contact between two surfaces and when a body has a relative motion with respect to a contact surface. In order to reduce friction the contact surface must be lubricated, also the friction depends on the coefficient of friction between surfaces and the normal force exerted by the surface parallel to the area of contact with the body. Mathematically it can be expressed with the following equation.
![F_{f} = u*N\\where:\\u = friction coefficient\\N = normal force [Newtons]\\F_{f}= friction force [Newtons]](https://tex.z-dn.net/?f=F_%7Bf%7D%20%3D%20u%2AN%5C%5Cwhere%3A%5C%5Cu%20%3D%20friction%20coefficient%5C%5CN%20%3D%20normal%20force%20%5BNewtons%5D%5C%5CF_%7Bf%7D%3D%20friction%20force%20%5BNewtons%5D)
Answer:
Power output = 96.506 watts
Explanation:
Drag coefficient (Cd) = 0.9
V = 7.3 m/s
Air density (ρ) = 1.225 kg/m^(3)
Area (A) = 0.45 m^2
Let's find the drag force ;
Fd=(1/2)(Cd)(ρ)(A)(v^(2))
So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N
Drag power = Drag Force x Drag velocity.
Thus drag power, = 13.22 x 7.3 = 96.506 watts
