Which of the the following distance vs time graphs represents an object the is moving at constant non zero velocity

A device known as Atwood's machine consists of two masses hanging from the ends of a vertical rope that passes over a pulley. As

Vlad [161]

Answer:

$a= \frac{(m_{1} -m_{2} )*g}{m_{1} +m_{2} }$

$T= \frac{m_{2}g(1+m_{1} -m_{2} ) }{m_{1}+m_{2} }$

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

M1 free body diagram : Look at the attached graphic

∑F = m₁*a

W₁ -T= m₁*a

W₁ - m₁*a = T Equation 1

M2 free body diagram :Look at the attached graphic

∑F = m₂*a

T-W₂= m₂*a

W₂ + m₂*a = T Equation 2

Equation 1 = Equation 2

W₁ - m₁*a = W₂ + m₂*a

W₁ - W₂ = m₁*a + m₂*a

m₁*g -m₂*g = a* (m₁ + m₂)

a = (m₁*g -m₂*g) / (m₁ + m₂)

$a= \frac{(m_{1} -m_{2} )*g}{m_{1} +m_{2} }$

Calculation of the tension in the rope (T)

We replace a in the equation 2

W₂ + m₂*a = T

W₂ + m₂*g*(m₁ -m₂) / (m₁ + m₂) = T

m₂*g + m₂*g*(m₁ -m₂) / (m₁ + m₂) = T

m₂*g( (1+ (m₁ -m₂)) / (m₁ + m₂) = T

m₂*g (1+m₁ -m₂) / (m₁ + m₂) = T

$T= \frac{m_{2}g(1+m_{1} -m_{2} ) }{m_{1}+m_{2} }$

7 0

3 years ago

How high is the highest waterfall

LenKa [72]

Angel Falls is 3,230 feet, which is the tallest waterfall in the world.
Hope this helps!
Happy studying,
~Mistermistyeyed.

5 0

3 years ago

An element can be identified by _____.

Darina [25.2K]

Answer:

The answer is the number of protons in its atom

Explanation:

5 0

3 years ago

Read 2 more answers

Two thin parallel conducting plates are placed 2.0 cm apart. Each plate is 2.0 cm on a side; one plate carries a net charge of 8

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

The total electric field.

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

E = 2.26 10⁶ N/C

In conclusion, using Gauss's law and that charges of the same sign repel each other, we can find the result for the questions about the charged plates:

The charge inside the plates is zero.

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

Learn more about Gauss's law here: brainly.com/question/15175106

6 0

3 years ago

The _______ is the layer of the atmosphere in which weather occurs

seraphim [82]

The "Troposphere" <span>is the layer of the atmosphere in which weather occurs!

It is the lowermost layer of atmosphere, where we live.

Hope this helps!</span>

8 0

3 years ago