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3 years ago

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The Tevatron accelerator at Fermilab (Illinois) is designed to carry an 11 mA beam of protons traveling at very nearly the speed

of light (300,000,000 m/s) around a ring 6300 m in circumference. How many protons are stored in the beam

1 answer:

Lunna [17]3 years ago

3 0

Answer:

The number is $N= 1.442*10^{12} \ protons$

Explanation:

From then question we are told that

The current carried is $I_p = 11mA =11 *10^{-3}A$

The speed is $c = 3.0*10^8 m/s$

The circumference is $D = 6300 \ m$

Mathematically current is expressed as

$I = \frac{Q}{t}$ where Q is the quantity of charge passing through a conductor and t is the time in seconds

The time t can be mathematically represented as

$t = \frac{D}{ c}$

Substituting this into the equation fro current

$I = \frac{Q}{\frac{D}{c} }$

Now making the quantity of charge the subject of the formula

$Q = \frac{I * D}{c}$

Generally the number of proton in the beam is mathematically represented as

$N = \frac{Q}{n}$

Where n is the charge on one proton with a value of $n= 1.602*10^{-19}C$

Now substituting for q in the equation for N

$N = \frac{I * D}{c * n}$

Substituting values

$N = \frac{11*10^{-3} * 6300}{ 3.0*10^8 * 1.602 *10^{-19}}$

$N= 1.442*10^{12} \ protons$

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Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci

Alecsey [184]

Given Information:

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:

Energy produced = P = ?

Answer:

P = 2.03 J/s

Explanation:

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

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3 years ago

What is a 7 letter word that describes the relationship between frequency and period?

adoni [48]

Present. NOT SURE IF THIS IS EVEN RIGHT!! I JUST GUESSED AND THOUGHT

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To impress his friends while riding on a carnival

s344n2d4d5 [400]

Answer:

B. decreases while his angular speed remains unchanged.

Explanation:

His angular speed will always be the same as the wheel's angular speed, which remains constant as it's in uniform motion. As for linear speed, which is defined as the product of angular speed and distance r to the center of rotation, and his distance to center is decreasing, his linear speed must be decreasing as well.

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A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic

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Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

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3 years ago

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

Margarita [4]

Answer:

A) $h = 69.58 m$

D) $v = 58.12 \frac{m}{s}$ (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

$V_{ox} = V_{o}Cos\alpha _{o}$ Formula (1)

$V_{oy} = V_{o}Sin\alpha _{o}$ Formula (2)

Where:

Vo: Initial velocity in m/s

$\alpha_{o}$ : Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

$v_{y}^{2} = v_{oy}^{2} -2gy$ Formula (3)

Where:

$v_{f}^{2}$ : Final speed component in vertical direction in m/s

$v_{oy}^{2}$ : Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

$v_{x} = v_{ox}$ Because the movement is uniform in the x direction (constant speed)

$v_{y} = v_{oy}-g*t$ Formula (4) Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

$v_{o} = 65.2 \frac{m}{s}$

$\alpha _{o}$ = 34.5º above the horizontal

$g=9.8 \frac{m}{s^{2}}$

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

$v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}$

$v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}$

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

$0=(36.93)^2-2*9.8*h$

$h=\frac{ (36.93)^2}{2*9.8} = 69.58 m$

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

$v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}$

$v_{x} = v_{ox} = 53.7 \frac{m}{s}$

$v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}$ (Speed magnitude)

$\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})$

α = 22.49° (Speed direction above the horizontal)

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3 years ago