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Shalnov [3]
2 years ago
13

1. A Faraday ice pail similar to the one you will use in the experiment is used in this question. The principle of its operation

relies on the proportionality between the charge on the inner pail and the potential difference between the inner pail and ground. Suppose the relationship between charge q and potential difference � is given by � = � ∙ 4.30×10JZ . When a charged wand is inserted into the inner pail and touches it, the electrometer reading is −16.0 �����. (a) How much charge resided on the wand? (b) Before grounding the pail, a second wand is inserted into the ice pail and touches it. The electrometer reading now is +28.0 �����. How much charge resided on the second wand? (c) If the two wands were separated by 7.50 �� before they were discharged onto the ice pail, what magnitude of force did either exert on the other? Was it attractive or repulsive? Assume that no leakage of charge occurs during the experiment and that charge on the wand is same as the charge on the pail. Recall that J ^_`a = 9.00×10c ��e �e .
Physics
1 answer:
algol [13]2 years ago
8 0

Answer:

your mommy

Explanation:

YOUR MAMMA

YOUR MAMMA

I LOVE MY DADDY

HE DOES ME REAL HARD

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A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
2 years ago
An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which
Irina-Kira [14]

Answer:

2\times 10^{-3}\ C

6000

1.2 J

3.33\times 10^{-6}\ F

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C

The charge flowing through the electrocytes in that amount of time is 2\times 10^{-3}\ C

The maximum potential is given by

V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000

The number of electrolytes is 6000

Energy is given by

E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F

The equivalent capacitance of all the electrocyte cells in the electric eel is 3.33\times 10^{-6}\ F

8 0
2 years ago
A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the en
Nina [5.8K]

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

         Em₀ = U = m g h₁

Final point. Lower (slide bottom)

           Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

            mgh₁ = ½ m v² + mgh₂

            v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

          x = v₀ₓ t

         y =  t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

        t = √ 2h₂ / g

We substitute in the other equation

        d = √ (2g (h₁-h₂))  √ 2h₂ / g

        d = √ (4 h₂ (h₁-h₂))

        H = h₁ + h₂

        h₁ = H -h₂

        d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

7 0
2 years ago
Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor
grin007 [14]

The job that the fan is designed and built to do is to convert the electrical energy it uses into the kinetic (motion) energy of moving air.

I can't really guarantee that it accomplishes that with MOST of the electrical energy it uses, because I don't know how efficient your fan is. For example, if it's a really old fan, and one blade has the end broken off, and a lot of dust and mosquitoes have gotten into the motor, and it shakes and vibrates and makes a lot of noise when it's running, then it's converting a lot of the electrical energy into thermal energy (it gets hot when it runs) and some into sound energy too.

If you can live without the word "most" in the question, then we can assume that the fan is well designed and running like a top, and the answer is definitely choice-B .

4 0
2 years ago
To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?
koban [17]

<em>To determine the y component of velocity of a projectile </em><u><em>sine </em></u><em>operation is performed on the angle of launch.</em>

<u>Answer:</u> <em>sine</em>

<u>Explanation:</u>

Thus a_x=0,a_y=g

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion v= u+at

Thus velocity along x direction v_x=u cos θ

Velocity along y direction v_y= u sinθ -gt

Sign of g is negative.

3 0
2 years ago
Read 2 more answers
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