When was the international space station launched

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecul

pogonyaev

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.

For more details regarding hydrogen bonding, visit:

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5 0

1 year ago

A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

m_a_m_a [10]

Answer:

F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

$F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N$

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0

3 years ago

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

2 years ago

Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maxi

meriva

Answer:

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

Explanation:

So we are told that it is a RC circuit. We are told $Q = C V [1 - e^(-t/RC)]$ = 12.0 V, R = 1.07 MΩ and C = 2.66 µF.

a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F

τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

τ = 2.85 s

b.) The relationship between capacitance, potential, charge is given:

$Q = CV[1-e^{-t/RC} ]$

The capacitor is fully charge when t approaches infinity, therefore:

$Q = \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]$

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

$Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}$

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

$(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s$

t = 1.691 s

Hope this helps! I'm not sure what the units you want, so convert to the desired units.

6 0

3 years ago

Convert 3.2 kilometers to feet?

Tanzania [10]

Answer:

10498.7 feet

Explanation:

7 0

2 years ago

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