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3 years ago

Which of the following are density labels? a. Kg/L b. g/m c. g/mL d. cm/g

Physics

2 answers:

fomenos3 years ago

6 0

Answer: a. kg/L and c. g/mL

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of object is measured in units of grams and kilograms.

Volume of object is measured in units of $(length)^3$ and are represented as ml, L, $m^3$ , $cm^3$

Thus density can be represented in units kg/L and g/ml

g/m is mass per unit length.

cm/g is length per unit mass.

bearhunter [10]3 years ago

4 0

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance.<span> </span><span>It has standard units of kg/m^3 or g/mL. So, the best answer is option C.</span>

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A spaceship is traveling toward Earth while giving off a constant radio signal with a wavelength of 1 meter (m). What will the s

Nadya [2.5K]

Answer:

Less than 1 m

Explanation:

When objects are getting closer to each other there is a slight change in the wavelength that is being transmitted by either objects. This is known as the blue shift of waves. Here, the wavelength reduces.

In the opposite case the when objects are getting farther from each other there is a slight change in the wavelength that is being transmitted by either objects. This is known as the red shift. Here, the wavelength increases.

In this case the spaceship is getting close to Earth hence the wavelength will be lower than 1 m.

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3 years ago

A wheel rotates at 24 revolutions every 3 minutes. This is equivalent to

mrs_skeptik [129]

8 revolutions every minute

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2 years ago

a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?

evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

$\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}$

So,

$h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}$

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

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yan [13]

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3 years ago

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ANTONII [103]

Answer:

it is important to create measurable goals bc :

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hope it helps!

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3 years ago

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