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kati45 [8]
2 years ago
7

In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and

determine how each one affects how quickly the cart moves along the track. In the space below, write a scientific question that you can answer by doing this experiment.

Physics
2 answers:
Veseljchak [2.6K]2 years ago
7 0

Answer:

Sample Response: How do force and mass affect the acceleration of an object?

Explanation:

Savatey [412]2 years ago
3 0

<u>Answer:</u>

<em>Newtons II law: </em>

<em>     </em>It  is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

<em>For the hanging weight/mass</em>

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

           Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

        Now, Fnet = m₁ .g - T

                          = m₁.a

So From Newtons II law<em> F =  m.a</em>

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EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
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C) (KE + PE)beginning = (KE + PE) end.
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D) All of the above.
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4 0
2 years ago
Read 2 more answers
To move a suitcase up to the check-in stand at the airport a student pushes with a horizontal force through a distance of 0.95 m
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Answer:

33.68 N

Explanation:

Data

W= 32J

d- 0.95m

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W=Fd

They are asking for the magnitude which is the force, so you need to solve for force.

F=W/d

= 32J/ 0.95m

= 33.68 N

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2 years ago
An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
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Find the given attachment for solution

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Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a p
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B = 0.046T

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size of the screen = 51.2cm

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3 0
2 years ago
A 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what i
never [62]

Answer:

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p)  -

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We know that ,

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Let m_{1} and m_{2} be the mass of the person and the book respectively and v_{1} and v_{2} be the final velocities of the person and book respectively.

∴p_{f}=m_{1}v_{1}+m_{2}v_{2}=0

From the question ,

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v_{2} = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × v_{1} )+ (2.44×8.25) = 0

∴v_{1}  = - 0.2687 m/s (Negative sign suggests that the motion of  the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

4 0
2 years ago
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