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brilliants [131]
3 years ago
7

In an electricity demonstration at the Deutsches Museum in Munich, Germany, a person sits inside a metal sphere of radius 0.90 m

. Charge is applied to the sphere until the magnitude of the electric field on the outside surface of the sphere is 3.0×105N/C. When the demonstration is over, the person climbs out unharmed. During the demonstration, what is the magnitude of the electric field inside the sphere?
Physics
1 answer:
vladimir2022 [97]3 years ago
6 0

Answer:

 E_interior = 0

Explanation:

As the sphere is metallic, the electrical charges are distributed on its surface, as far away as possible from each other.

If we apply Gauss's law, as the charge is on the surface, when drawing a spherical Gaussian surface, we see that there is no charge inside, therefore there is no electric field inside the metallic sphere.

          E_interior = 0

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A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

W=mg: weight of the ladder, with m = 20 kg (mass) and g=9.8 m/s^2 (acceleration of gravity)

W_M=Mg: weight of the person, with M = 54 kg (mass)

N_1: normal reaction exerted by the wall on the ladder

N_2: normal reaction exerted by the floor on the ladder

F_f = \mu N_2: force of friction between the floor and the ladder, with \mu (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}

where

Wsin 21^{\circ} is the component of the weight of the ladder perpendicular to the ladder

W_M sin 21^{\circ} is the component of the weight of the man perpendicular to the ladder

N_1 sin 69^{\circ} is the component of the normal  force perpendicular to the ladder

And solving for N_1, we find the force exerted by the wall on the ladder:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of N_2.

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

\sum F_y = 0\\N_2 - W - W_M =0

And substituting and solving for N2, we find:

N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

\sum F_x = 0\\F_f - N_1 = 0

And re-writing the equation,

\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

\mu = \frac{N_1}{N_2}

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)

And substituting, we get

N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0
3 years ago
DNA is coiled into chromosomes in a cell's
tia_tia [17]

Answer: Cell's nucleus

Explanation :

The full form of DNA is Deoxyribonucleic acid. The genetic information of a cell is organized in the DNA. It is inherited from parents by their children.

DNA is coiled into chromosomes in a cell's nucleus. It has a shape like a double helix. It is twisted in the form of spirals.

Hence, DNA is coiled into chromosomes in a cell's Nucleus.

6 0
3 years ago
Read 2 more answers
Please please help
dsp73

Answer:

true

Explanation:

The law of conservation of charge states that whenever electrons are transferred between objects, the total charge remains the same.

3 0
3 years ago
Which statement correctly describes the current in a circuit that is made up of any two resistors connected in parallel with a b
AysviL [449]
B is the answer I think
8 0
3 years ago
A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
Montano1993 [528]

Net Force = (mass) x (acceleration)  (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

6 0
2 years ago
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