Answer:
Net Force = 10N
Acceleration = 2m/s^2
Explanation:
calculate the net force and the acceleration on the block
Net force on the block F = mass * acceleration
Net force acting in the positive direction = 4N + 6N = 10N
Mass = 5kg
According to newton's second law;
a = F/m
a = 10N/5
a = 2m/s^2
hence the acceleration on the block is 2m/s^2
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Hey There!</h2><h2>
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Answer:</h2><h2 /><h2>

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<h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
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SOLUTION:</h2>
According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,

W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.



<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>
I think the correct answer from the choices listed above is option D. The model of the universe that suggests that the sun is the center of the universe was first brought by Copernicus. His model is known as the "Sun centered model".
Answer:
Acceleration a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Explanation:
For the truck to accelerate without losing its load.
Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.
Fa ≤ F(friction)
But;
Fa = mass × acceleration
Fa = ma
ma ≤ F(friction)
a ≤ (F(friction))/m ......1
Given;
Fa = mass × acceleration
Fa = ma
mass m = 800 kg
F(friction) = 2400 N
Substituting the given values into equation 1;
a ≤ F(friction)/m
a ≤ 2400N/800kg
a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2