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Airida [17]
3 years ago
10

A neutral atom of an element has the same number of __________ and ________.

Physics
2 answers:
scoray [572]3 years ago
8 0
The answer is protons and neutrons (C)
Volgvan3 years ago
7 0

The answer is protons and neutrons (C)

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The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
Nana76 [90]

Answer:

E

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice  the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

7 0
3 years ago
What is the magnitude of the speed of a person at the equator due to rotational speed of the Earth?
notka56 [123]
The circumference of the Earth at the equator is listed as 24,901 miles.
So his speed is

                     24,901 miles per day.

Convert it to units that we have a better feel for:

                   (24,901 mi/da) x (1 da / 24 hrs)

             =    (24,901 / 24)  (miles/hour)

             =  about  1,038 miles per hour.

You'll find a huge number of people on the internet these days,
telling you that you could not be moving at that speed and not
feel it, so therefore the Earth is not spinning, and it's not a globe.

I have a lot of feelings and comments about those people, their
lines of reasoning, and their levels of education and intelligence,
so don't get me started.

I just want to guarantee you that everything you're learning about
the Earth and the solar system in school is well founded, and it's
all based on the life's work of some of the smartest people of the
past 300 years of human history.  Everything you're taught about
the Earth has good reasons behind it, whereas those other people
have nothing.

A person on Earth's equator is moving from west to east at roughly
1,038 miles per hour, relative to any point on the Earth's rotation axis.
4 0
3 years ago
Directions: Using the T-chart below, compare balanced forces and unbalanced forces.
Salsk061 [2.6K]

Explanation:

unbalanced: a turning vehicle, apple falling on the ground, kicking a ball

balanced: floating on water, fruit hanging from tree, tug of war equally balanced teams

8 0
2 years ago
A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has
AlladinOne [14]

Given Information:

Angular displacement = θ = 51 cm = 0.51  m

Radius = 1.8 cm = 0.018 m

Initial angular velocity = ω₁ = 0 m/s

Angular acceleration = α = 10 rad/s ²

Required Information:

Final angular velocity = ω₂ = ?

Answer:

Final angular velocity = ω₂ = 21.6 rad/s

Explanation:

We know from the equations of kinematics,

ω₂² = ω₁² + 2αθ

Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.

ω₂² = (0)² + 2αθ

ω₂² = 2αθ

ω₂ = √(2αθ)

We know that the relation between angular displacement and arc length is given by

s = rθ

θ = s/r

θ = 0.51/0.018

θ = 23.33 radians

finally, final angular velocity is

ω₂ = √(2αθ)

ω₂ = √(2*10*23.33)

ω₂ = 21.6 rad/s

Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.

3 0
3 years ago
An air bubble of volume 20 cm³ is at the bottom of a lake 40 m deep, where the temperature is 4.0°C. The bubble rises to the sur
soldi70 [24.7K]

Answer:

100 cm³

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

n and R are constant, so:

P₁V₁/T₁ = P₂V₂/T₂

If we say point 1 is at 40m depth and point 2 is at the surface:

P₂ = 1.013×10⁵ Pa

T₂ = 20°C + 273.15 = 293.15 K

P₁ = ρgh + P₂

P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa

P₁ = 4.933×10⁵ Pa

T₁ = 4.0°C + 273.15 = 277.15 K

V₁ = 20 cm³

Plugging in:

(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)

V₂ = 103 cm³

Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.

6 0
3 years ago
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