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1 year ago

What is the difference between a mineral, glass, and a rock?

Physics

1 answer:

IrinaVladis [17]1 year ago

3 0

Answer:

A mineral is a solid formation that occurs in earth. A rock is a combination of more than just one mineral. And glass have more of a crystalline structure

Explanation:

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A white dwarf is:______.

bija089 [108]

Answer:

the exposed core of a dead star, supported by electron degeneracy pressure.

Explanation:

A white dwarf is a low luminosity exposed core of a dead star having mass comparable to the sun but volume comparable to the earth . So its density is very high . These stars have lost the capacity to generate energy through the process of fusion . Due to high gravitational energy , it goes on shrinking but ultimately balanced by electron degeneracy pressure. It is not a main sequence star as it has lost the power of fusion .

6 0

2 years ago

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

Likurg_2 [28]

Answer:

k = 4422.35 KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k

k=Spring constant

So we can say that

( 2 π f)² = m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³ k

33167.69 = 7.5 x 10⁻³ k

k=4422.35 x 10³ N/m

k = 4422.35 KN/m

Therefore spring constant will be 4422.35 KN/m

4 0

2 years ago

URGENT <br><br> Give three examples of models that you see every day

umka2103 [35]

Models are very common. The ingredients list on a bottle of ketchup is a model of its contents, and margarine is a model of butter. A box score from a baseball game is a model of the actual event. A trial over an automobile accident is a model of the actual accident.

7 0

1 year ago

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

$E = \frac{1}{2}kx^{2}$ .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

$8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}$

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

$E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}$

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0

2 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

3 years ago