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3 years ago

An electron is a particle with a ____.

Physics

2 answers:

Dahasolnce [82]3 years ago

4 0

A beta particle. Hoped I help. Sorry if it wrong.

Mkey [24]3 years ago

3 0

An electron is a particle with a<em> negative electrical charge of 1.6 x 10⁻¹⁹ Coulomb </em>and <em>about 9.1 x 10⁻³¹ kg of mass</em>.

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The maximum height a typical person can jump is about 60cm (0.6m). By how much does the gravitational potential energy increase

sasho [114]

The gravitational potential energy will increase by 423.36 J

<h3>How to determine the potential energy at ground level</h3>

Mass (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 0 m
Potential energy at ground level (PE₁) =?

PE = mgh

PE₁ = 72 × 9.8 × 0

PE₁ = 0 J

<h3>How to determine the potential energy at 60 cm (0.6 m)</h3>

Mass (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 0.6 m
Potential energy at 60 cm (0.6 m) (PE₂) =?

PE = mgh

PE₂ = 72 × 9.8 × 0.6

PE₂= 423.36 J

<h3>How to determine the change in potential energy </h3>

Potential energy at ground level (PE₁) = 0 J
Potential energy at 60 cm (0.6 m) (PE₂) = 423.36 J
Change in potential energy =?

Change in potential energy = PE₂ - PE₁

Change in potential energy = 423.36 - 0

Change in potential energy = 423.36 J

Learn more about energy:

brainly.com/question/10703928

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1 year ago

Consider a sample containing 1.70 mol of an ideal diatomic gas.

babunello [35]

I don't know

because I don't know

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2 years ago

At a race car driving event, a staff member notices that the skid marks left by the race car are

krok68 [10]

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

$v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s$

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

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5 0

2 years ago

Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rot

riadik2000 [5.3K]

To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as

$a_c = \omega^2 r$

Here,

$\omega =$ Angular velocity

r = Radius

Our values are given as,

$\omega = 4.12*10^{16}rad/s$

$r = 5.29*10^{-11}$

Replacing,

$a_c = (4.12*10^{16})^2( 5.29*10^{-11})$

$a_c = 8.979*10^{22}m/s^2$

Therefore the electron's centripetal acceleration is $8.979*10^{22}m/s^2$

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3 years ago

Density is calculated by dividing

Yanka [14]

Density is defines as the ratio of mass to volume.

So you measure the mass and volume of a sample, and
divide the mass by the volume, to find the density.

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3 years ago