Answer:
Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.
Answer:
<em>(C) If the composition of a mixture appears uniform no matter where you sample it, is homogeneous; sand on a beach *IS HETEROGENEOUS* because when you look at it up close, you can identify different types of particles, such as sand, shells, and organic matter.</em>
Explanation:
<em>(A) Pure Water is a collection of solely H2O molecules therefore Pure Water is classified as a *Compound*.</em>
<em>(B) Table Salt is NOT a heterogeneous mixture because the particles of salt can't be separated, and it is a *Pure Substance*.</em>
<em>(D) Maple Syrup is a homogeneous mixture because the solutes are fully dissolved and not easily identified. In other words, Maple Syrup is uniform throughout.</em>
<em>-Hope this helps!</em>
<em />
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
The correct answer for this question is this one: "C. Neither Natalie nor Will." Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the people in the household. <span>Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age.</span>
(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
#SPJ4
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