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3 years ago

Which of the following would most likely happen if water did not form hydrogen bonds? (2 points)

1 answer:

5 0

The answer choices for this question are:

A) Water would exist only as a solid.

B) Water would be able to dissolve all ionic compounds.

C) Water would not expand when it freezes.

D) Water would not exist on Earth.

The right answer is the option C) Water would not expand when if freezes.

Explanation:

Expansion of water when it freezes is an atypical behavior because substances contract when they freeze with the correpondant increase in density.

But when water forms the ice crystals, the strong hydrogen bonds do not permit an arrangement in which the molecules of water are closer and so they occupy more space (more volume). The length of the hydrogen bonds limits the distance between each water molecule. So, these bonds are the responsible of the expansion of water as it freezes.

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Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that

kaheart [24]

Answer:

Young's modulus for the rope material is 20.8 MPa.

Explanation:

The Young's modulus is given by:

$E = \frac{FL_{0}}{A\Delta L}$

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

$A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}$

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

$F = T_{w} = W_{o}$

Where:

$T_{w}$ : is the tension of the wire

$W_{o}$ : is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

$F = mg = 1700 kg*9.81 m/s^{2} = 16677 N$

Hence, the Young's modulus is:

$E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa$

Therefore, Young's modulus for the rope material is 20.8 MPa.

I hope it helps you!

7 0

3 years ago

An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha

hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

$F_D =F_g$

By definition we know that the Drag force is defined as

$F_D= \frac{1}{2} C_d \rho A V^2$

Where,

$C_d =$ Drag coefficient

$\rho =$ Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

$F_g = (m_1 +m_2) g$

Where,

$m_1 =$ Mass of sphere

$m_2 =$ Mass of unknown object

Equating the two equations we have to

$(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2$

Re-arrange for m_2,

$m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1$

Our values are given by,

$C_d = 0.5$

$\rho = 1.22Kg/m^3$

$V = 66.7m/s$

$m_1 = 3Kg$

$d= 32.7*10^{-2}m$

$r = 16.35*10^{-2}m$

Replacing in the equation we have,

$m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3$

$m_2 = 8.62Kg$

Therefore the mass of unknown object is 8.62Kg

4 0

3 years ago

If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?

REY [17]

Answer:

The car's angular speed is $\frac{rad}{s}$ .

Explanation:

Angular velocity is usually measured with $\frac{radians}{sec}$ , so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω) is given by:

$V = \omega R$

Using the values from the question, we get:

$2 \frac{m}{s} = \omega (0.4m)$

$\omega = 5 \frac{rad}{s}$

Therefore, the car's angular speed is $5 \frac{rad}{s}$ .

Hope this helped!

3 0

3 years ago

What a not an aspect of relative wind

daser333 [38]

Relative wind is defined as the airflow relative to an airfoil.

4 0

3 years ago

Read 2 more answers

A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po

saul85 [17]

The location of the point F that partitions a line segment from D to E ( $\overline{DE}$ ), that goes from negative 4 to positive 5, into a 5:6 ratio is fifteen halves (option 4).