Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:


From Eqn(2), we see that

so using Eqn(3) on Eqn(1), we get

Solving for the acceleration, we see that


b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

Since the crate started from rest,
Thus our equation reduces to



Answer:
C.) 1.5 kg
Explanation:
Start with the equation:

Plug in what you know, and solve:

Find matching soluation:
C.) 1.5 kg
<span> A. stops accelerating </span>
<span> B. stops speeding up </span>
<span> C. experiences equal forces of gravity and air resistance </span>
<span> D. all of the above</span>
A full lake is dropping at its constant rate, meaning it is unchanging over time.
After 4 weeks’ time, it already dropped 3 feet.
Now, let’s solve for the Unit rate based on the given data.
=> 3 feet in 4 weeks
=> Unit rate is 3 feet per 4 weeks
=> in 4 weeks there are (7 * 4) = 28 days
=> 3 feet / 28 days
=> the Unit rate is .107 feet / day
Answer:
Wl = 1740 N
Explanation:
maximum lift weight unaided = force exerted (F) = 650 N
length of the wheelbarrow (L) = 1.4 m
weight of the wheelbarrow (w) = 80 N
distance of center of gravity of the wheel barrow from the wheel = 0.5 m
distance of center of gravity of the load from the wheel = 0.5 m
find the weight of the load (Wl)
from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive
ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0
(F x 1.4) = ((Wl x 0.5) + (w x 0.5)
Wl =
Wl =
Wl = 1740 N
Using lens equation;
1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)
Substituting;
1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm
Therefore, the object should be place 99.23 cm from the lens.