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2 years ago

What is the radius of an automobile tire that turns with a period of 0.091 s and has a linear

Physics

1 answer:

faust18 [17]2 years ago

3 0

Answer:

1) The radius of the tire is approximately 0.28966 meters

2) The centripetal force is the force that keeps a body moving on a circular path

Explanation:

1) The linear speed of the automobile tire = 20.0 m/s

The period with which the tire turns = 0.091 s

The period = The time it takes to make a complete turn

Therefore;

The number of turns in 1 second = 1/0.091 ≈ 10.989 turns

The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire

∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire

From the speed of the car, 20.0 m/s, we have;

The distance covered in 1 second = 20.0 meters

Therefore;

10.989 × 2 × π × Radius of the tire = 20.0 meters

Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters

The radius of the tire ≈ 0.28966 meters

2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.

The centripetal force is given by the formula, $F = \dfrac{m \cdot v^2}{r}$

Where;

F = The centripetal force

m = The mass of the object

v = The linear velocity of the object

r = The radius of the rotational motion.

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Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

$\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L$

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0

2 years ago

A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106

defon

Answer:

$4.7\cdot 10^{-14}N$

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

$F=qvB$

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

$q=1.6\cdot 10^{-19} C$

$v=6.5\cdot 10^6 m/s$

$B=4.5\cdot 10^{-2} T$

So the force experienced by the electrons is

$F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N$

3 0

2 years ago

Find the intensity in decibels [i(db)] for each value of i. normal conversation: i = 106i0 i(db) = power saw a 3 feet: i = 1011i

White raven [17]

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

8 0

2 years ago

Read 2 more answers

A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

Lena [83]

Answer:

E = k Q 1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

λ= Q / x = dq / dx

dq = λ dx

We substitute in the equation

∫ dE = k ∫ λ dx / x²

We integrate

E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

E = k Q 1 / (x₀-x₂) (x₀-x₁)

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2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)