Answer:
a. 10.2 N b. 1.73 c. 58.8 N d. 10.2 N
Explanation:
a. The force that will make the block slide down the plane
This is the component of the block's weight along the plane F = mgsinθ where m = mass of block = 1.2 kg, g = 9.8 m/s² and θ = angle of incline = 60°.
So, F = mgsinθ = 1.2 kg × 9.8 m/s² × sin60°.= 10.18 N ≅ 10.2 N
b. The coefficient of friction
The coefficient of friction, μ = tanθ = tan60° = 1.73
c. The normal reaction.
This is equal to the vertical component of the block's weight. So, F = mgcosθ. Substituting the values for the variables from above, we have
F = mgcosθ = 1.2 kg × 9.8 m/s² × cos60°.= 58.8 N
d. The frictional force
Since the block does not slide, there is no net force on it. If f is the frictional force, then
F - f = ma. Since a = acceleration = 0,
F - f = 0
f = F = mgsinθ = 1.2 kg × 9.8 m/s² × sin60°.= 10.18 N ≅ 10.2 N
b. biosphere
c. asthenosphere
d. magnetosphere
The box has 3 forces acting on it:
• its own weight (magnitude w, pointing downward)
• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)
• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ F = -f + w sin(35°) = m a
• net perpendicular force:
∑ F = n - w cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
n = w cos(35°)
n = (15 kg) (9.8 m/s²) cos(35°)
n ≈ 120 N
Solve for the mag. of friction:
f = µn
f = 0.25 (120 N)
f ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a
a ≈ (54.3157 N) / (15 kg)
a ≈ 3.6 m/s²
Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:
v² - v₀² = 2 a ∆x
v² = 2 (3.6 m/s²) (3 m)
v = √(21.7263 m²/s²)
v ≈ 4.7 m/s
B. the rate of evaporation of a fluid
C. how much energy it takes to warm a fluid
D. the density of a fluid
E. a fluid's resistance to flowing
Answer:
E. a fluid's resistance to flowing
Explanation:
Viscosity, also known as "thickness" is a rheological property that describes a fluid's resistance to flowing, fluids of low viscosity, like water, flow more easily while high viscosity fluids, like mud, are harder to move through. It is an important property because it determines the energy required to make a certain fluid flow.