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natita [175]
3 years ago
12

An object with a mass of 5.0 kg accelerates 8.0 m/s² when an uknown force is applied to it. what is the amount of the force?​

Physics
2 answers:
Harman [31]3 years ago
5 0
  • Mass=5kg
  • Acceleration=8m/s^2

Using newtons second law

\\ \sf\longmapsto Force=Mass\times Acceleration

\\ \sf\longmapsto Force=5(8)

\\ \sf\longmapsto Force=40N

maks197457 [2]3 years ago
4 0

<h2>We know </h2>

The second law of Newton that

<h2>Force=Mass×Acceleration</h2>

Given ,

  • Mass=5kg

• \: acceleration \:  = 8 \frac{m}{s}^{2}

<h3>Therefore Using newtons second law</h3>

\begin{gathered}  \: Force=Mass\times Acceleration\end{gathered}

\begin{gathered} \\ \sf\  = Force=5 \times 8\end{gathered}

so \: force \:  = 40n

Answer - 40 N

Hope it helps you!!

#IndianMurga

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A block of mass
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(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

6 0
2 years ago
Caculate the component of a force of 200 ns <br>at a direction of 60° to the force​
Lady bird [3.3K]

Answer:

F_x = 100N

F_y = 100\sqrt 3 \ N

Explanation:

Given

F = 200N

\theta = 60^o

Required

The component of the force in F direction

To do this, we simply calculate the force in the vertical and horizontal direction.

This is calculated as:

F_x = F * \cos(\theta) --- Horizontal

F_y = F * \cos(\theta) ---- Vertical

So, we have:

F_x = F * \cos(\theta) --- Horizontal

F_x = 200N * \cos(60^o)

F_x = 200N * 0.5

F_x = 100N

F_y = F * \cos(\theta) ---- Vertical

F_y = 200N * \sin(60^o)

F_y = 200N * \frac{\sqrt 3}{2}

F_y = 100\sqrt 3 \ N

7 0
3 years ago
Which statement best describes the work and energy in these examples?
german

Answer:

the văd and food are exemple of energy

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4 years ago
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irinina [24]

Answer:

Diffusion

Explanation:

Diffusion is the tendency of molecules to rise from region of higher concentration to region lower concentration until they are even distributed. The tendency of molecules to move toward areas of lower concentration is called diffusion. Diffusion occurs base on diffusion gradient which is change in quantity of the the molecule with respect to change in variables of the medium. The rate of diffusion varies for different substances and mediums.

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