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Leto [7]
3 years ago
6

Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We

will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field between the plates points upward.a) An electric field is established by applying a potential difference to the plates. It is found that a field of strength E0 will cause the droplet to be suspended motionless. Write an expression for the droplet's charge q. Let g be the acceleration due to gravity. Express your answer in terms of the suspending field E0 and the droplet's weight mg.
Physics
1 answer:
Katen [24]3 years ago
4 0

Answer:

q=\frac{mg}{E_o}              

Explanation:

Given:

Charge = <em>q</em>

Electric field strength =E_o

weight of the droplet = <em>mg</em>

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, F=qE_o

This is balanced by the weight, mg

Equating the two:

qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}

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makvit [3.9K]

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

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2. Now remote sends signal to Television, this is a long range interaction.

7 0
3 years ago
Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a
nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

    p₀o = m₁ v₁ + m₂ v₂

    pf = 0

    m₁ v₁ + m₂ v₂ = 0

    m₁ / m₂ = -v₂ / v₁

    m₁ / m₂=  - (-6.2) / 4.7

     m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

    Xcm = 1/M    (m₁ x₁ + m₂ x₂)

We derive from time

   Vcm = 1/M   (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

    0 = 1/M   (m₁ v₁ + m₂v₂)

   m₁ v₁ + m₂v₂ = 0

    m₁ / m₂ = -v₂ / v₁

   m₁ / m₂ = 1.3

4 0
3 years ago
Reading Check: (Describe) how pressure changes as the velocity of a fluid<br> increases.
juin [17]

Answer:

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Explanation:

4 0
3 years ago
On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is
Alchen [17]
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback

So for conservation of momentum,
rho = mv

M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf

For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2

Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
6 0
3 years ago
Please help me:(
Lelechka [254]

Answer:B

Explanation:

I’m doing the same thing you are!

3 0
3 years ago
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