Yes, it is. Physical science<span>, the systematic study of the inorganic world</span>
Answer:
YFy = 0 = Ffsinθ + Fncosθ - Fw
Explanation:
From the base of the vector Fn, draw a vertical line. the small angle between this line and Fn is also theta. The component of Fn in the vertical direction is Fncos(theta).
Take a moment to picture extreme cases. Sine is 0 at 0 and 1 at 90. Cosine is 1 at 0 and 0 at 90.
Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.
We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:
Vf² = Vi² + 2ad
Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.
Given values:
Vi = 0m/s (dumbbell starts falling from rest)
a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)
d = 80×10⁻²m
Plug in the values and solve for Vf:
Vf² = 2(10)(80×10⁻²)
Vf = ±4m/s
Reject the negative root.
Vf = 4m/s
The momentum of the dumbbell is given by:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
m = 10kg
v = 4m/s (from previous calculation)
Plug in the values and solve for p:
p = 10(4)
p = 40kg×m/s
Answer:
Total length of spring 0.647 m
Explanation:
We have given mass of the person m = 150 kg
Acceleration due to gravity 
Spring constant k = 10000 N/m
Nominal length of spring = 0.50
According to hook's law
x = 0.147 m
So total length of spring = 0.50+0.147 = 0.647 m
Work = (force) x (distance)
(550 newtons) x (0.5 meter) = 275 joules each lift .
(275 joules/lift) x (10 lifts) = 2,750 joules of work all together.
Power = (work done) / (time to do the work)
= (2,750 joules) / (20 seconds)
= 137.5 watts . (about 0.18 horsepower)
