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IgorLugansk [536]
3 years ago
13

A 10 kg computer accelerates at a rate of 5 m/s2. How much force was applied to the computer?​

Physics
2 answers:
worty [1.4K]3 years ago
8 0

Answer:

<h2>50 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 10 × 5

We have the final answer as

<h3>50 N</h3>

Hope this helps you

AURORKA [14]3 years ago
6 0

Answer:

<h2><u>Given </u><u>:</u><u>-</u><u> </u></h2>
  • Mass = 10 kg
  • Acceleration = 5 m/s²
<h2><u>To </u><u>find</u></h2>

Force applied

<h2>SoluTion</h2>

As we know that Force applied is the product of mass and acceleration

\sf \: force \:  = 5 \times 10

\sf \: force \: =  50 \: n

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There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
2 years ago
A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is
Maru [420]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q  = -5.02×10^{-9}C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8×10^{-2}m) (21.8×10^{-2}m)  = 4.75×10^{-4}m²

A  = 4.75×10^{-4}m²

Surface charge density = Q/A

Surface charge density = (-5.02×10^{-9}C)/(4.75×10^{-4}m²)

Surface charge density = -1.057×10^{-5} Cm^{-2}

3 0
3 years ago
What is power?
Blizzard [7]

Answer:

D

Explanation:

P=Work/Time

The rate at which work is done matches that.

6 0
3 years ago
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A
vazorg [7]

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

8 0
3 years ago
Please help!!
IgorLugansk [536]

Answer:

Si un objeto se mueve en relación a un marco de referencia (por ejemplo, si una profesora se mueve a la derecha con respecto al pizarrón, o un pasajero se mueve hacia la parte trasera de un avión), entonces la posición del objeto cambia. A este cambio en la posición se le conoce como desplazamiento. La palabra desplazamiento implica que un objeto se movió, o se desplazó.

Explanation:

El desplazamiento se define como el cambio en la posición de un objeto. Se puede definir de manera matemática con la siguiente ecuación:

\text{desplazamiento}=\Delta x=x_f-x_0desplazamiento=Δx=x  

f

​  

−x  

0

​  

start text, d, e, s, p, l, a, z, a, m, i, e, n, t, o, end text, equals, delta, x, equals, x, start subscript, f, end subscript, minus, x, start subscript, 0, end subscript

x_fx  

f

​  

x, start subscript, f, end subscript se refiere al valor de la posición final.

x_0x  

0

​  

x, start subscript, 0, end subscript se refiere al valor de la posición inicial.

\Delta xΔxdelta, x es el símbolo que se usa para representar el desplazamiento.

Debemos ser cuidados al usar la palabra distancia, ya que hay dos maneras de usar el término en física. Podemos hablar acerca de la distancia entre dos puntos, o podemos hablar de la distancia recorrida por un objeto.

La distancia se define como la magnitud o el tamaño del desplazamiento entre dos posiciones. Observa que la distancia entre dos posiciones no es la misma que la distancia recorrida entre ellas.

Es importante darse cuenta que la distancia recorrida no tiene que ser igual a la magnitud del desplazamiento (es decir, la distancia entre dos puntos). De manera específica, si un objeto cambia de dirección en su trayecto, la distancia total recorrida será mayor que la magnitud del desplazamiento entre esos dos puntos. Ve los ejemplos resueltos a continuación.

8 0
3 years ago
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