Answer:
hydorgen bond
Explanation: pls give me branliest
That's 105 km that he flew, or 65.2 miles ! I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking. But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:
The displacement is the distance and direction from the place
where the crow took off to the place where he landed.
-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.
D² = (60 km)² + (45 km)²
= 3,600 km² + 2,025 km² = 5,625 km²
D = √(5625 km²) = 75 km .
-- It's direction is the angle whose tangent is (45 S / 60 W).
tan⁻¹ (45/60) = tan⁻¹ (0.75) = 36.9° south of west
= 53.1° west of south.
= not exactly southwest but close.
1. If I hadn't bought that car yesterday.
2. I would have never bought the tickets for today..
3. She wouldn't have gotten suspended.
4. We could've worked on our project.
5. but I was in the shower.
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Answer:
I think it's 2 the photo is hard to tell what they are exactly talking about.
