Answer:
The answer to your question should be D.
Explanation:
reactants are on the laft side of arrow and products are on right side of arrow
Answer:
p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J
Explanation:
The dipole moment of a dipole is the product of charges by distance
p = 2 a q
With 2a the distance between the charges and the magnitude of the charges
p = 1.7 10⁻⁹ 6.8 10⁻⁶
p = 1.16 10⁻¹⁴ C m
The potential energie dipole is described by the expression
U = - p E cos θ
Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line
Orientation parallel to the field
θ = 0º
U = 1.16 10⁻¹⁴ 1160 cos 0
U1 = 1.35 10⁻¹¹ J
Antiparallel orientation
θ = 180º
cos 180 = -1
U2 = -1.35 10⁻¹¹ J
The difference in energy between these two configurations is the subtraction of the energies
ΔU = | U1 -U2 |
ΔU = 1.35 10-11 - (-1.35 10-11)
ΔU = 2.7 10 -11 J
Answer:
(1) A sound wave a mechanical wave because mechanical waves rely on particle interaction to transport their energy, they cannot travel through regions of space that are void of particles. Sound is a mechanical wave and cannot travel through a vacuum. These particle-to-particle, mechanical vibrations of sound conductance qualify sound waves as mechanical waves. Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave. The answer is(B) it travels in the medium.
(2) An ocean wave is an example of a mechanical transverse wave
The compression is the part of the compressional wave where the particles are crowded together. The rarefaction is the part of the compressional wave where the particles are spread apart. The answer is (C) Compression.
It kinda makes sense that it would be considered a metamorphosis.
Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
