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2 years ago

What is the acceleration of a 100 kg object pushed by Beau with a force of 500 N? Write out your equation.

Physics

1 answer:

Aneli [31]2 years ago

4 0

Answer:

ProvideD :

Mass ( m ) = 100 kg , Force ( F ) = 500 N

To FinD :

Acceleration ( a )

SolutioN :

We know , F = ma so put up the values and then solve for a

$\longrightarrow \tt \: 500 = 100a$

$\longrightarrow \tt \: \frac{500}{100} = a$

$\longrightarrow \tt \: a = \: \boxed { \tt{5 \: ms ^{ - 2} }}$

The acceleration produced is 5 ms ^ -2 .

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How many electrons are in 204 C of charge?

zubka84 [21]

Answer:

The mass number 204 – 82 protons = 122 neutrons

Explanation:

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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3 years ago

Calculate the gravitational potential energy is a bicycle of mass 20kg and its rider of mass 50kg cycle to the top of a hill 120

Travka [436]

Explanation:

Gravitational Potential Energy can be calculated with the following formula:

$mgh$

Where m is mass, g is Gravitational Field Strength, and h is height. GFS on Earth is always 9.81, the combined mass of the cyclist and the bicycle is 70, and the height is 120. Multiplying these values together, we get:

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1 year ago

an optician uses a plane mirror to help him. suppse a patient sits in a chair 2.5m away from him. He views the image of a chart

viva [34]

Answer:

I think 75 m

Explanation:

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3 years ago

Um automóvel percorre 6,0 km para o norte e, em seguida 8,0 km para o leste. A intensidade do vetor posição, em relação ao ponto

Valentin [98]

Answer:

Explanation:

A car travels 6.0 km to the north and then 8.0 km to the east. The intensity of the position vector, in relation to the starting point is: a) 14 km b) 2.0 km c) 12 km d) 10 km e) 8.0 km

Check attachment for diagram

The intensity of the position vector is equal to the displacement,

So, to calculate the displacement, we need to find the length of the straight line from starting point to end point.

So, applying Pythagorean theorem

c² = a² + b²

R² = 6² + 36²

R² = 36 + 64

R² = 100

R = √100

R = 10 km.

Verifique el adjunto para ver el diagrama

La intensidad del vector de posición es igual al desplazamiento,

Entonces, para calcular el desplazamiento, necesitamos encontrar la longitud de la línea recta desde el punto inicial hasta el punto final.

Entonces, aplicando el teorema de Pitágoras

c² = a² + b²

R² = 6² + 36²

R² = 36 + 64

R² = 100

R = √100

R = 10 km.

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3 years ago