Answer:
32 m and -2.4 m/s
Explanation:
Given:
v₀ = 25 m/s
t = 2.8 s
a = -9.8 m/s²
Find: Δy, v
Δy = v₀ t + ½ at²
Δy = (25 m/s) (2.8 s) + ½ (-9.8 m/s²) (2.8 s)²
Δy = 31.6 m
v = at + v₀
v = (-9.8 m/s²) (2.8 s) + 25 m/s
v = -2.44 m/s
Rounded to two significant figures, the bullet reaches a height of 32 m and a velocity of -2.4 m/s.
Answer:
the time at which it passes through the equilibrum position is:
t = 0.1 second
Explanation:
given
w= 4pounds
k(spring constant) = 2lb/ft
g(gravitational constant) = 10m/s² = 32ft/s²
β(initial point above equilibrum) = 1
velocity = 14ft/s
attached is an image showing the calculations, because some of the parameters aren't convenient to type.
Answer:
pretty sure its B if it isnt im so so sorry
Explanation:
The answer is C
The colour that you see is what is being reflected and the colour that you don't see is what is being absorbed. In this case, you don't see any colour, so all wavelengths are being absorbed.
It is a kinetic type of jump