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-Dominant- [34]
3 years ago
9

Describe what a homogeneous electric field is, and where we find such

Physics
1 answer:
slega [8]3 years ago
7 0

Answer:

no one really answers on here anymore but can u click on my profile and help me with my question

Explanation:

the answer is A homogeneous electric field has the same magnitude and direction at any place. A good example of a homogeneous field is the field between two charged metal plates. The field strength depends on the voltage U and the plate distance d.

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Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
Bob is pushing a box across the floor at a constant speed of 1.5 m/s, applying a horizontal force whose magnitude is 60 n. alice
earnstyle [38]

120n

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7 0
4 years ago
Determine qual é a carga elétrica de um corpo que possui 1 milhão de partículas?
MaRussiya [10]

Answer:

This can be translated to:

"find the electrical charge of a body that has 1 million of particles".

First, it will depend on the charge of the particles.

If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C

Then the total charge of the body will be:

Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C

If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)

Q = 1.6*10^-13 C

But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.

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