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baherus [9]
3 years ago
5

A simple elevator ride can teach you quite a bit about the normal force as this rider below can (hopefully) tell you. There are

three different scenarios given, detailing the rider\'s experience in an unnamed hotel. For each scenario, calculate the normal force, FN,1-3, acting on the rider if his mass is m = 76.6 kg and the acceleration due to gravity g = 9.81 m/s2. In scenario 1, the elevator has constant velocity. In scenario 2 the elevator is moving with upward acceleration a2 = 4.84 m/s2. Finally, in scenario 3, unfortunately for the rider, the cable breaks and the elevator accelerates downward at a3 = 9.81 m/s2.
FN1= ___________
FN2= ___________
FN3= ___________-
Physics
1 answer:
miskamm [114]3 years ago
4 0

Answer:

FN1 = 751.5 N

FN2 = 1122.2 N

FN3 = 0

Explanation:

Scenario 1 :

  • The elevator has constant velocity.

The normal force, can adopt any value, as needed by Newton's 2nd Law, in order to fit this general expression:

Fnet = m*a

In the first  scenario, as the elevator is moving at a constant speed, this means that no external net force is present.

The two forces that act on the rider, are gravity (always present, downward) and the normal force, as follows:

Fnet = Fn - m*g = m*a

For scenario 1:

Fnet = 0 ⇒ Fn = m*g = 76.6 kg * 9.81 m/s² = 751. 5 N

  • Scenario 2

In this scenario, the elevator has an upward acceleration of 4.84 m/s², so the Newton's 2nd Law is as follows:

Fnet = FN - m*g = m*a  

⇒ FN = m* ( g+ a) = 76.6 kg* (9.81 m/s² + 4.84 m/s²) = 1,122.2 N

  • Scenario 3

As the elevator is in free fall, this means that a = -g, so, in this condition, the normal force is just zero, as it can be seen from the following equation:

FN-mg = m*a

If a = -g,

⇒ FN -mg = -mg ⇒ FN=0

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Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o
VikaD [51]

Answer:

For the point inside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

We have to integrate it over the ring, which is an angular integration.

E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

6 0
3 years ago
A sample of gas has an initial volume of 4.5 L at a pressure of 754 mmHg . Part A If the volume of the gas is increased to 8.5 L
Firdavs [7]

Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

6 0
4 years ago
A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c
Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

Q=mC_s \Delta T

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

\Delta T=40.0 C - 80.0 C=-40.0 C is the change in temperature of the wire

Substituting into the equation, we find

Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J

And the sign is negative because the heat is released by the wire.

6 0
3 years ago
A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
miskamm [114]

Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

net force at highest point

F_{net}=W+F_c

where F_c= centripetal force

F_c=\dfrac{mv^2}{r}

F_{net}=F_{n}= Normal Force

585=652+F_c

F_c=-67\ N

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

F_n=F_c+W

F_n=652+67

F_n=719\ N  

8 0
4 years ago
What are non-clastic?
igor_vitrenko [27]
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8 0
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