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3 years ago

The density of 2 kilograms of iron on Earth's surface is

Physics

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

B. The same on the moon.

Explanation:

The density of an object is the ratio of the mass contained by the object to the volume occupied by that mass.

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

When the object is taken from the earth to anywhere in the universe, its mass remains constant. The dimensions of the object and hence its volume also remains constant anywhere in the universe.

Therefore, the density of the object will also remain the same as it depends upon the mass and the volume of the object.

So, the correct option is:

B. The same on the moon.

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A wave is travelling at 3000 m/s has a wavelength of 1 m.

Levart [38]

Answer:

a] 3000hz

b]3.33 × 10⁻⁴

ci]300

ii] 3000

iii]60,000

Explanation:

3 0

3 years ago

What happens to the light rays when they hit the specimen?

pogonyaev

Answer:

Explanation:

Ray of light when hits any specimen or object. The light is partially reflected, partially reflected and partially absorbed. It is never completed reflected, refracted or absorbed. Hence, the correct answer would be c.

4 0

3 years ago

PLEASE HELP ME,, I WOULD BE SO HAPPY

Juliette [100K]

Answer:

Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(

Explanation:

6 0

3 years ago

Read 2 more answers

Many biological systems are well-described by the laws of statistical physics. A simple yet often powerful approach is to think

GuDViN [60]

Answer:

z1/z2

Explanation:

we have no quantum effects therefore we can make use of Maxwell Boltzmann distribution in the description of this system.

using the boltzman distribution the probability of finding a particle in energy state

$P_{ei} = \frac{gie^{-ei/kol} }{z}$

we have

gi to be degeneration of the ith state

ei to be energy of ith state

$z=e^{-ei/kbt}$ summation

$P_{ope} = \frac{e^{-ei/kBt} }{z} = \frac{Z_{1} }{Z}$

We have R to be equal to

$\frac{P_{ope} }{P_{Close} } = \frac{Z1}{Z2}$

8 0

3 years ago

The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?

Greeley [361]

On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where
I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore 
I = 1/12m*2² = 1/3m kg*m² 

The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so 
ω = 2π * 2 rev/s = 4π s^(-1) 

The angular momentum would therefore be 
L = Iω 
= 1/3m * 4π 
= 4/3πm kg*m²/s, where m is the rod's mass in kg. 

The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. 

Edit: 650 g = 0.650 kg, so 
L = 4/3π(0.650) kg*m²/s 
≈ 2.72 kg*m²/s

4 0

3 years ago