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3 years ago

7

An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10 µC that does not touch the

conductor. There are no other charges in the vicinity. Which statement about this conductor is true?

1 answer:

aliina [53]3 years ago

8 0

Answer:

Explanation:

we have to make charge inside the conductor zero because we know that electric field inside the conductor should be zero

so, the outer surface of the conductor should contain + 10 uC of charge and the inner surface contains -10 uC

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What is the function of hammer and anvil in human ear

MArishka [77]

Answer:

Vibrations of the eardrum are in turn translated into oscillations of tiny bones (ossicles) found within the middle ear. The Hammer, Anvil, and Stirrup magnify the oscillations and relay this information to the chamber of the inner ear.

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3 years ago

Read 2 more answers

A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$

8 0

3 years ago

A stationary charge is located between the poles of a horseshoe magnet. Is a magnetic force excerted of the charge? why?

vredina [299]

A stationary charge is located between the poles of a horseshoe magnet. The magnetic force exerted by the charge is zero.

<h3>What is charge?</h3>

Charge is the physical property of matter which cause a particle to attract or repel when placed in its field.

A stationary charged particle does not interact with a static magnetic field. A charge placed in a magnetic field experiences a magnetic force. There will be no magnetic force acting on a stationary charge. The charge must be moving in order to have magnetic force on it.

Thus, the magnetic force exerted by the charge is zero.

Learn more about charge.

brainly.com/question/19886264

#SPJ4

7 0

2 years ago

Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets

PolarNik [594]

Step by step solution :

standard deviation is given by :

$\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}$

where, $\sigma$ is standard deviation

$\bar{x}$ is mean of given data

n is number of observations

From the above data, $\bar{x}=24.88$

Now, if $x=24.8$ , then $(x-\bar{x})^2=0.0064$

If $x=23.9$ , then $(x-\bar{x})^2=0.9604$

if $x=26.1$ , then $(x-\bar{x})^2=1.4884$

If $x=25.1$ , then $(x-\bar{x})^2=0.0484$

If $x=24.5$ , then $(x-\bar{x})^2=0.1444$

so, $\sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}$

$\sum (x-\bar{x})^2 =2.648$

$\sqrt{\sum \frac{(x-\bar{x})^2}{n}}$

$\sigma =0.7277$

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

6 0

3 years ago

Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) $x=\frac{1}{2}at^2+v_0t+x_0$

The equation for velocity v and a constant acceleration a is:

(ii) $v=at+v_0$

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) $a = \frac{v -v_0}{t}$

(iv) $x = \frac{v-v_0}{2t}t^2+v_0t + x_0$

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) $x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t$

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v: $v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}$

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

$f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}$

5 0

3 years ago