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3 years ago

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(science) If an 50 kg object moves at 3 m/s2, what is the object’s force? Use the equation above, show your work, and include un

its.

2 answers:

Sliva [168]3 years ago

6 0

$\huge\mathfrak{\fcolorbox{aqua}{lime}{Answer}}$

Given :

$Mass\:\: is\: \:50\: \:kg \:$

$Acceleration\: = \:{3m/s}^{2}$

while finding force make sure that mass is in (Kg) and acceleration is in (m/s²)

$Force\:=\:ma$

$Force\:=\:50kg\:×\:{3m/s}^{2}$

$Force\:=\:150 \:N$

bekas [8.4K]3 years ago

3 0

Answer:

$\displaystyle F = 150 \ N$

General Formulas and Concepts:

<u>Dynamics</u>

Newton's Second Law of Motion: $\displaystyle \sum F = ma$

Explanation:

We are given the following variables:

m = 50 kg

a = 3 m/s²

To find the object's force, we use Newton's Second Law of Motion. Plug in your variables into the equation:

F = (50 kg)(3 m/s²)

F = 150 kg · m/s²

F = 150 N

Topic: AP Physics 1 - Algebra Based

Unit: Dynamics

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F= ma (formula)

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azamat

Gauss's law and charges of the same sign repel allows us to find the results for the questions about the charged plates are:

The charge inside the plates is zero.

The field in the middle of the plates is: E = 2.26 10⁶ N/C

<h3>Gauss's law.</h3>

Gauss's law says that the electric flux through a Gaussian surface is proportional to the charge inside it.

Ф = ∫ E . dA = $\frac{q_{int}}{\epsilon_o }$

where Ф is the flux, E the electric field, A the area, $q_{int}$ the charge inside the surface.

They indicate that we have two metallic plates with a charge of 80 μC = 80 10⁻⁶ C in each one, since the plate is metallic, the electrons are free to move in it and repel each other, therefore the ones that are farthest from each other are placed, this is concentrated on the surface of the metal plate, therefore the charge inside the surface is zero.

Let's use Gauss's law to find the electric field, we define a Gaussian surface with a cylinder base parallel to the plate, in this case the field created by the charge is parallel to the normal of the surface of the plates.

2 E A = $\frac{q_{int}}{\epsilon_o}$

The two comes from the fact that the electric field is emitted towards both sides of the plate.

The charge density on each plate is:

σ = q A

Let's substitute.

E A = $\frac{\sigma A}{2 \ \epsilon_o}$

The electric field is a vector magnitude, so vector addition must be used, see attached for the direction of the electric field.

$R_{total} = E_1+E_2$

$E_{total} = \frac{\sigma}{\epsilon_o}$

Let's calculate.

The charge density.

$\sigma = \frac{q}{l^2}$

$\sigma = \frac{ 80 \ 10^{-6} } { 2.0 \ 2.0}$

σ = 20 10⁻⁶ C

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E = $\frac{20 \ 10^{-6} }{8.85 \ 10^{-12} }$

E = 2.26 10⁶ N/C

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The charge inside the plates is zero.

The field in the middle of the plates is: E = 2.26 10⁶ N/C

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Apparent magnitude:

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