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3 years ago

What stops gravity from pulling you to the center of the earth

Physics

2 answers:

Artyom0805 [142]3 years ago

8 0

Answer:

electron degeneracy pressure...or in other word electro magnetic repulsion

zmey [24]3 years ago

8 0

The ground and everything under it do.

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Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

$E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s$

The velocity of the flea when leaving the ground is 1.327363 m/s

$W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m$

The flea will travel 0.00090243 m upward

8 0

3 years ago

When Missourians carve their initials into the bark of a tree, the damage leaves the tree open to ____________________.

sasho [114]

Answer:

In fact, carving letters into a tree probably won't hurt it. ... In general, the tree will compartmentalize the wound and it will heal over. The initials that remain visible are essentially scar tissue, permanent scar tissue.

Explanation:

Unfortunately, when carving into the trunk of a tree the blade of a knife often penetrates the outer bark and cuts into the inner bark. ... In cases that the phloem is damaged all the way around the trunk (in a ring for example), the tree will slowly and eventually starve to death.

add my s n a p

luke_raines19

5 0

3 years ago

You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds

german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

$f_{beat}=\frac{1}{T}$

Then the frequency for the previous period given (2sec) is

$f_{beat}=\frac{1}{2}$

$f_{beat} = 0.5Hz$

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

$f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz$

Hence option A is incorrect.

We can do this process for 254Hz as $f_1$ and 258 Hz for $f_2$ , then

$f_{beat} =|254Hz-258Hz|$

$f_{beat} = 4Hz$

Hence option B is incorrect.

We can also do this process for 255Hz as $f_1$ and 257 Hz for $f_2$ , then

$f_{beat} =|255Hz-257Hz|$

$f_{beat} = 2Hz$

Hence option C is incorrect.

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

$f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz$

Hence option D is incorrect.

We can also do this process for 255.75Hz as $f_1$ and 256.25 Hz for $f_2$ , then

$f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz$

Hence option E is incorrect.

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0

3 years ago

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to

Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0

3 years ago

Which of the following are true statements? A. Like charges repel B. Unlike charges repel C. Unlike charges attract or D. Charge

Brums [2.3K]

Answer: The correct answers are (A) and (C).

Explanation:

The expression from electrostatic force is as follows;

$F=\frac{kq_{1}q_{1}}{r^{2} }$

Here, F is the electrostatic force, k is constant, r is the distance between the charges and $q_{1},q_{1}$ are the charges.

The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.

Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.

The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.

Therefore, the true statements from the given options are as follows;

Like charges repel.

Unlike charges attract.

7 0

3 years ago