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3 years ago

An element's atomic number is 64. How many protons would an atom of this element have?

Physics

2 answers:

yuradex [85]3 years ago

6 0

It would have 64 protons

Gwar [14]3 years ago

3 0

It would have 64 protons :)

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The diagram shows the electric field near conducting sphere a which of the following statements is correct

Ludmilka [50]

I really don’t know but I hope this helps

4 0

3 years ago

Why is it important to lean forward and listen

galben [10]

To show your engaged in the learning topic and to show your undivided attention

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3 years ago

Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π

Kruka [31]

Answer:

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

w = $\sqrt{\frac{g}{L} }$

angular velocity and period are related

w = 2π / T

we substitute

2π / T = \sqrt{\frac{g}{L} }

T = $2\pi \ \sqrt{\frac{L}{g} }$

In this exercise indicate that for a long Lo the period is To, then and increase the long

L = L₀ + ΔL

we substitute

T = $2\pi \ \sqrt{\frac{L + \Delta L}{g} }$

T = $2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }$

in general the length increments are small ΔL/L «1, let's use a series expansion

$\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...$

we keep the linear term, let's substitute

T = $2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )$

if we do

T = T₀ + ΔT

T₀ + ΔT = $2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}$

T₀ + ΔT = T₀ + $\pi \sqrt{\frac{1}{Lg} } \ \Delta L$

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

4 0

3 years ago

Two sacks contain the same number of identical apples and are separated by a distance r. The two sacks exert a gravitational for

stepan [7]

Answer:

Explanation:

im sleppy

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3 years ago

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0

2 years ago