Answer:
(A.)Nuclear fission and beta decay (electron emission)
<span>Data:
mass =
110-g bullet
d = 0.636 m
Force =
13500 + 11000x - 25750x^2, newtons.
a) Work, W
W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =
W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =
W = 8602.6 joule
b) x= 1.02 m
</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02
W = 10383.5
c) %
[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.
</span>
Answer:
Explanation:
Electric field strength= Force/unit charge
E= (kQq/r²)/q ₓ r
where r is the unit vector in the direction of unit charge
E=
Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s
Is it not standard deviation? or am i just dumb lol
