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Bas_tet [7]
3 years ago
6

In the laboratory, a ball is dropped onto a force-sensing platform several times, each time hitting a different surface (foam, f

eathers, clay, etc.). The momentum of the ball changes by the same amount in each trial; in each trial, the average scale reading is F, and the time of collision t are measured. What quantities would need to be graphed to exhibit a straight-line relationship
Physics
1 answer:
NeX [460]3 years ago
5 0

Answer:

Graphing the momentum against the change in moment yields a linear relationship.

Explanation:

This is an impulse experiment,

          I = ∫ F .dt

where the force and time of the collision are measured, therefore if we assume an average force the integral reduces to

           I = F t

Furthermore, the momentum is equal to the change in moment of the ball, this change in moment can be found using the energy relations measuring the height of the ball and calculating its speed, in the two intervals for the descent and for the exit, possibly the heights are different so the moment change is different from zero.

Starting point. Higher

          Em₀ = U = mgh

Lower end point, just before hitting the scale

          Em_{f} = K = ½ m v²

in the path in the air there is no friction

          Em₀ = Em_{f}

          m g h = ½ m v²

          v = \sqrt{2gh}

this height is different for the descent and ascent of the ball, so we have two moments

         Δp = p_{f} - p₀

         Δp = m (v_{f} -v₀)

         

therefore we have the relationship

         

         I = Δp

Graphing the momentum against the change in moment yields a linear relationship.

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Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
Newtons first law stateS that object will move With a constant velocity if nothing acts on it. Does our every day experience con
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Answer:

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Explanation:

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4 0
3 years ago
A 3 kg object is moving along a horizontal surface. The kinetic energy of the object is increasing at a constant rate of 6 J/m;
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To solve this problem we will apply the concepts of energy conservation and Newton's second law that defines force as the product of the object's mass with its acceleration. Additionally we will apply concepts related to the kinematics equations of linear motion.

For conservation of energy we have that work is equal to kinetic energy therefore,

W = KE

Fd = \frac{1}{2} mv^2

Here,

F = Force

d = Displacement

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v= Velocity

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F = \frac{W}{d}

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7 0
4 years ago
The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N
Umnica [9.8K]

As per given conditions there are two directions along which forces are acting

1. Net force along left direction is given as

F_{left} = 352N + 176 N = 528 N

2. Net force towards right direction is given as

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now since the two forces here in opposite direction so here we will have net force given as

F_{net} = F_{right} - F_{left}

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F_{net} = 440 N

so here net forces must be 440 N towards right

7 0
4 years ago
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