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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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The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
2 years ago
You are using a rope to lift a 14.5 kg crate of fruit. Initially you are lifting the crate at 0.500 m/s. You then increase the t
lina2011 [118]

Answer:

W = 172.5 J

Explanation:

given,                                    

mass of the fruit crate = 14.5 kg

initial velocity to lift = 0.500 m/s

increase in the tension = 150 N

lift of crate = 1.15 m                  

work done by the tension = ?        

work done  = force x displacement

W = F s cos θ                                

θ = 0°                                    

W = F s x cos 0                                  

W = 150 x 1.15 x 1                

W = 172.5 J                                      

Work done on the crate by the tension force = W = 172.5 J

5 0
2 years ago
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Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
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Combine all of the x's on one side of the equation and then finish the problem!
8 0
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