Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. The given statement is true.
<h3>What are hydrogen bonds?</h3>
A hydrogen bond is an electrostatic force of attraction among a hydrogen atom tightly attached to a more electronegative "donor" atom or group and another electronegative atom bearing a lone pair of electrons, known as the hydrogen bond acceptor.
Hydrogen bonds are too flimsy to connect atoms to form molecules, but they do hold various portions of a single large molecule together in a specific three-dimensional shape.
Thus, the given statement is true.
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Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
Answer:
Fission is the opposite of fusion and releases energy only when heavy nuclei are split. As noted in Fusion, energy is released if the products of a nuclear reaction have a greater binding energy per nucleon than the parent nuclei.
The amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only 0.1 percent of the original nuclei is converted to energy. Daughter nucleus, energy, and particles such as neutrons are released as a result of the reaction
Answer:
17.2 seconds
Explanation:
Given:
v₀ = 0 m/s
a₁ = 10.0 m/s²
t₁ = 3.0 s
a₂ = 16 m/s²
t₂ = 5.0 s
a₃ = -12 m/s²
v₃ = 0 m/s
Find: t
First, find v₁:
v₁ = a₁t₁ + v₀
v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)
v₁ = 30 m/s
Next, find v₂:
v₂ = a₂t₂ + v₁
v₂ = (16 m/s²) (5.0 s) + (30 m/s)
v₂ = 110 m/s
Finally, find t₃:
v₃ = a₃t₃ + v₂
(0 m/s) = (-12 m/s²) t₃ + (110 m/s)
t₃ = 9.2 s
The total time is:
t = t₁ + t₂ + t₃
t = 3.0 s + 5.0 s + 9.2 s
t = 17.2 s
Round as needed.
Answer: 704
Explanation:Vi = 0 m/s
vf = 65 m/s
a = 3 m/s2
d = ??
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m 2/m2)/(6 m/s2) = d
d = 704 m
