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3 years ago

Covalent bonds are characterized by...

Physics

2 answers:

tresset_1 [31]3 years ago

5 0

Answer:

a.) Sharing of electrons

Explanation:

A covalent bond is a form of chemical bonding which is characterized by the sharing of electrons between atoms.

Alik [6]3 years ago

4 0

Answer:

A. the sharing of electrons

Explanation:

Covalent bonds share electrons.

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A force of 15N is exerted on an area of 0.1 m squared. what is the pressure

MrMuchimi

Answer:

1The force is with u. Lol no Idea if this is correct sry if u fail. I go must now

8 0

3 years ago

Lena is studying the properties of light in a laboratory. If she increases the amplitude of the light waves she is studying, wha

zaharov [31]

Answer:

A brighter light

Explanation:

Light waves travel through space via light particles called photons. This particles have in essence 2 properties: 1. Amplitude and 2.Frequency. The first one has to do with the intensity of light we see and the second one has to do with the energy (color). If we change only the amplitude, we will see a lighter or darker light and will keep the same color in all amplitude changes. But if we modify the frequency, the intensity will keep the same and the color changes as we move into the light spectrum.

Thus, increasing the amplitude, we will perceive a brigher light.

4 0

3 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$

$Y=(24cm/tan(32.12))$

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0

3 years ago

The measurement of an exoplanet radius is measured in units compared to

maksim [4K]

After an exoplanet has been identified using a given detection method, scientists attempt to identify the basic properties of the planet which can tell us what it might be made of, how hot it might be, whether or not it contains an atmosphere, how that atmosphere might behave, and finally, whether the planet may be suitable for life. It is often useful to first determine basic properties of the parent star (such as mass and distance from the Earth). This is then followed by the use of planetary detection methods to calculate planetary mass, radius, orbital radius, orbital period, and density. The density calculation will provide clues as to what the planet is made of and whether or not it contains a significant atmosphere. Mass and Distance of Parent Star The mass and distance of an exoplanet's parent star must often be calculated first, before certain measurements of the exoplanet can be made. For example, determining the star's distance is an important step in determining a star's mass (see below). Knowing the mass of a star then allows the mass of the planet to be measured, for example when using the Radial Velocity Method.

7 0

3 years ago

Beginning at time t = 0, a student exerts a horizontal force on a box of mass 30 kg, causing it to move at 1.2 m/s toward an ele

Alexandra [31]

Since the initial and the final velocity are just equal, it is implied that the acceleration is zero. This means that the net force acting on the body is also zero. The horizontal force should be equal to the force of friction. The force of friction is equal to the product of the coefficient and the normal force.

Ff = (coefficient of friction) x Fn

The normal force is equal to the object's weight if the surface is horizontal.

Ff = (0.20) x (30 kg) x (9.8 m/s²) = 58.8 N


Thus, the horizontal force exerted must be 58.8 N.

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6 0

3 years ago