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DiKsa [7]
3 years ago
14

Particle accelerators, such as the Large Hadron Collider, use magnetic fields to steer charged particles around a ring Consider

a proton ring with 36 identical bending magnets connected by straight segments. The protons move along a 1.0-m-long circular arc as they pass through each magnet. Part A What magnetic field strength is needed in each magnet to steer protons around the ring with a speed of 2.0 x 10 m/s? Assume that the field is uniform inside the magnet, zero outside. Express your answer with the appropriate units
Physics
1 answer:
coldgirl [10]3 years ago
6 0

Answer:

\beta= 3.49x10^{-8}T

Explanation:

The magnetic field can be find using the equation

m*v^2/r=q*v*\beta

You can cancel a element of v'

m*v/r=q*\beta

C=36*1m=2\pi*r

r=\frac{36}{2\pi } =5.7295m

Solve to magnetic field

\beta=\frac{m*v^2}{r*q}

The charge and mass of the proton are:

m_p=1.6x10^{27}kg, q_p=1.6x10^{-19}C

Replacing numeric

\beta=\frac{1.6x10^{-27}kg*2x10m/s}{1.6x10^{-19}C*5.73m}

\beta= 3.49x10^{-8}T

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There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s
Maslowich

Answer:

.a = 849.05 m / s²

Explanation

The centripetal acceleration is

            a = v² / r

     

Linear and angular velocity are related

          v = w r

Angular velocity and frequency are related by

        w = 2π f

Let's replace

        a = w² r

         a = 4π² f² r

Let's reduce to the SI system

       f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s

       .r = 10.3 cm = 0.103 m

Let's calculate

       a = 4π² 14.45²  0.103

       .a = 849.05 m / s²

8 0
3 years ago
12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
True or False:<br><br> It is safe to touch an electric current while soaking wet.
olganol [36]

Answer:

False

Explanation:

Water contains ions that can conduct electricity and if touched it can cause harm (aka electrocution).

8 0
3 years ago
What percentage of energy consumed in the united states is nonrenewable??
IgorLugansk [536]
It was estimated, according to the United States Energy Information Administration, that the United States still consumes almost 91% of non-renewable resources as its energy source. In addition, 32 % of it are natural gas, 28 % for petroleum and crude oil, and 21 % for coal power.
6 0
3 years ago
A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then
xeze [42]
The anwser is A or D
7 0
3 years ago
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