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.
B is the answer!!!!!!!!!!!!!
Answer:
0.3m
Explanation:
Given parameters;
Elastic energy = 20J
Spring constant = 445N/m
Unknown
Final extension of the spring = ?
Solution:
The elastic potential energy of a stretched spring can be determined using the expression below;
EP = 
k e²
k is the spring constant
e is the extension
Insert the parameters and solve;
20 = x 445 x e²
multiply those sides by 2;
2(20) = 445e²
40 = 445e²
e² = 
= 0.09
e = √0.09 = 0.3m
Answer: c opinion is something ppl just suggest but a theory needs proof before it is confirmed
It’s designed to protect an electrical circuit from damage caused by overcurrent, usually resulting from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.
That’s really just the basic purpose.
Happy to help!
~Brooke❤️
