Answer:
By 16.7% or 0.167 IPM
Explanation:
Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:




The force on the ship is more than a car
V: velocity of wave
f: frequency
L: wavelenght
v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m
Answer:
(a) The equivalent spring constant is 598.485 N/m
(b) The work done is 46.926 J
Explanation:
From Hooke's law of elasticity
K (spring constant) = F/e
F is the range of force exerted = 237 - 0 = 237 N
e is the extension of bowstring = 0.396 m
K = F/e = 237/0.396 = 598.485 N/m
Work done = 1/2 Fe = 1/2 × 237 × 0.396 = 46.926 J