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sdas [7]
3 years ago
5

Find the acceleration for a force of 2 N acting on 2 kg.

Physics
1 answer:
igomit [66]3 years ago
4 0

Answer:

The acceleration of a 2 kg box acted on by a net force of 2 N is solved by using Newton’s second law of motion in the form acceleration = F_net / mass. Since the mass of the box is 2 kg and the net force is 2 N, the acceleration of the 2 kg box is 1 m/s^2.

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How many position coordinates are required to identify the position of a system undergoing one-dimensional motion? What symbol s
gregori [183]
One-dimensional motion can be plotted through the Cartesian plane which has a coordinates of (x,y). These coordinates are the abscissa and ordinates. Since, there are two coordinates, the answer to the second item is two.

The symbol that can be used to identify systems position is (x,y). Since this is one dimensional motion, it is possible that one of the two coordinates becomes zero. 
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A pilot is flying a small airplane at 40 m/s in a circular path with a radius of 200 m. The pilot has a mass of 80.5 kg
g100num [7]
Try This Method(substitute your values)
Centripetal force = Mass X Velocity^2/ Radius 
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3 years ago
Explain the process that causes dew to form on blades of grass?
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<span>Heat is radiated, atmospheric moisture condenses at a rate greater than that at which it can evaporate, resulting in the formation of water droplets.</span>
7 0
3 years ago
Read 2 more answers
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer
boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0
3 years ago
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