2 years ago

If this atom has a balanced charge, how many protons would you expect to find in this atom

Physics

2 answers:

guapka [62]2 years ago

8 0

Then, number of protons would be equal to number of electrons.

Pachacha [2.7K]2 years ago

4 0

The same number as the number of electrons in the cloud.

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Who was the first scientist to question the idea that atoms were uncuttable

gizmo_the_mogwai [7]

Answer:

Democritus

Explanation:

He called these "uncuttable" pieces atomos. This is where the modern term atom comes from. Democritus first introduced the idea of the atom almost 2500 years ago.

5 0

2 years ago

An apple falls from a tree and hits the ground 9.98 m below. with what speed will it hit the ground

cupoosta [38]

The answer to this question is <span>13,537</span>

8 0

2 years ago

What kind of image is formed by a convex lens when the object is less than a focal length away from the object?

maw [93]

Bigger image and can get clearer

3 0

2 years ago

rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon

Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m$

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0

2 years ago

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

$\dfrac{I_{CM}}{I}=\dfrac{2}{7}$

$I_{CM}=\dfrac{2}{7}I$

the correct answer is option E

4 0

2 years ago