Can you please stop pasting this question, just go to his profile and ask him.
Answer:
25.6 m/s
Explanation:
Draw a free body diagram of the sled. There are two forces acting on the sled:
Normal force pushing perpendicular to the hill
Weight force pulling straight down
Take sum of the forces parallel to the hill:
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 38.0°)
a = 6.03 m/s²
Given:
v₀ = 0 m/s
a = 6.03 m/s²
t = 4.24 s
Find: v
v = at + v₀
v = (6.03 m/s²) (4.24 s) + (0 m/s)
v = 25.6 m/s
Answer:
C. 3.00 s
Explanation:
Given:
Δy = 1.80 m − 46.0 m = -44.2 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 3.00 s
Answer:
0.037 N/m
Explanation:
The web acts as a spring, so it obeys Hook's law:
(1)
where
F is the force exerted on the web
k is the spring constant
x is the stretching/compression of the web
In this problem, we have:
- The mass of the fly is 
- The force exerted on the web is the weight of the fly, so:
- The stretching of the web is
So if we solve eq.(1) for k, we find the spring constant:
Answer:
no Jake arm is high so the weight of an object become high due to potential energy.
so, option A is correct ) Jake did more work
