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Natalka [10]
2 years ago
8

The charge within a small volume dv is dq=ρdv. the integral of ρdv over a cylinder of length l is the total charge q=λl within t

he cylinder. use this fact to to determine the constant ρ0 in terms of λ and r. hint: let dv be a cylindrical shell of length l, radius r, and thickness dr. what is the volume of such a shell?

Physics
2 answers:
azamat2 years ago
7 0
The volume of the shell that you described would be:
dV=2L\pi r dr
Now we can rewrite the given integral:
\lambda L=\int\rho dV=L\rho\int2\pi r dr \\ \lambda L =L\rho \pi r^2\\ \rho=\frac{\lambda}{\pi r^2}
I have attached the picture explaining how we got the formula for the volume.
On the picture, I marked the rectangle. You can of this rectangle as the base, and the height would be the circumference of the cylinder.

Step2247 [10]2 years ago
6 0

Answer:

Given

dq=density*dv

q=lamda*I

Taking double integration

density=lamda/2*pi*r^2

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A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t
LenKa [72]

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

    F ’= fr - Wx = 0

    W = F ’L cos 0

    W = 0

4 0
2 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll
marissa [1.9K]

Answer:

h = 0.0362\,m

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

U_{g,A} + K_{A} = U_{g,B} + K_{B}

U_{g,B} - U_{g,A} = K_{A} - K_{B}

m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}

The moment of inertia of the coin is:

I = \frac{1}{2}\cdot m \cdot r^{2}

After some algebraic handling, an expression for the maximum vertical height is derived:

m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}

h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}

h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }

h = 0.0362\,m

3 0
2 years ago
When a surface is submerged in a fluid, the resultant pressure force on the body acts in what manner? Assume no shear forces are
Alex17521 [72]

Answer:

Perpendicular to the surface of the body.

Explanation:

When a surface is submerged in a fluid the resultant pressure force on the body acts perpendicular to the surface of the body. This is because fluids cannot withstand nor exert sideways forces. One could obtain this analytically considering that the origin of this force comes from the movement of the fluid molecules.

5 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
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