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anastassius [24]
2 years ago
12

An electron is accelerated from rest by an electric potential difference of 437 V, and then enters a region of uniform magnetic

field. The magnetic field is oriented perpendicular to the velocity with a magnitude of 42.8 mT. What will the radius of the electron's path be once it enters the magnetic field
Physics
1 answer:
Lelu [443]2 years ago
6 0

The radius of the electron's path is once it enters the magnetic field 3.16*10^-4 m.

The electric capacity difference, additionally called voltage, is the outside work needed to bring a rate from one region to some other area in an electric discipline. The electric powered capacity distinction is the change of potential electricity skilled by way of a test rate that has a fee of +1.

The distinction in ability among two points represents the paintings worried or the electricity released in the switch of a unit amount of power from one factor to the other.

Electric ability is the work executed in step with unit fee in bringing the fee from infinity to that factor against electrostatic pressure. In a conductor, electrons float simplest when there is a difference in electric stress at its ends. This is additionally known as the capacity distinction.

Learn more about the electric potential difference here brainly.com/question/14306881

#SPJ4

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Explanation:

Edothermic process is the absorbtion of heat.

It is the opposite of exothermic process which is the release of energy. In edothermic process thermal energy is absorbed from the surrounding

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Which of the following is an organic compound?
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At the end of the passage, Sarah says to Emma, “You’re not the only one with tricks up your sleeves.” Explain what Sarah means b
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Sarah plans to use trickery and cunning to get back at Emma. Sarah thinks she is smarter than Emma.

Explanation:

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6 0
3 years ago
Which combination of units can be used to express the magnetic field?
Zolol [24]

Answer:

The magnetic field unit in the International System is the tesla (T). A tesla is defined as the magnetic field that exerts a force of 1 N (newton) on a load of 1 C (coulomb) that moves at a speed of 1 m / s within the field and perpendicular to the field lines

Explanation:

Magnetic induction or magnetic flux density (B), is the magnetic flux that causes a diffusion charge in motion for each unit of normal area to the direction of the flow. It is also called the magnetic field strength.

The unit of magnetic flux density in the International System of Units is the tesla (T).

The tesla (symbol T), is the magnetic induction unit (or magnetic flux density) of the International System of Units (SI). It is defined as a uniform magnetic induction that, normally distributed over a surface of one square meter, produces through this surface a total magnetic flux of a weber.

<u>Equivalences: </u>

1 T = 1 Wb · m-2 = 1 kg · s-2 · A-1 = 1 kg · C-1 · s-1

A Tesla is also defined as the induction of a magnetic field that exerts a force of 1 N (newton) on a load of 1 C (coulomb) that moves at a speed of 1 m / s within the field and perpendicular to the lines of magnetic induction.

1 T = 1 N · s · m-1 · C-1

Basic Unit in the Cegesimal System of Units (CGS): Gauss (G)

A gauss (G) is a magnetic field unit of the Cegesimal System of Units (CGS). A gauss (G) is defined as a maxwell per square centimeter.

1 gauss = 1 maxwell / cm2

A gauss is equivalent to 10-4 tesla:

1 T = 10,000 G

8 0
4 years ago
A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i
AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume v and enthalpy h as,

h_1 = 95.96Btu/lb (  h_1 = h_f at 2psia )

v_1 = 0.016238ft^3/lb ( v_1 = v_f at 2psia )

w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb

w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb

h_3 = 1364.0Btu/lb

s_3 = 1.5073Btu/lb.R

( at P_3 = 1500psia & T_3 = 800^0F )

P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765

( S_f & S_{fg} when pressure is 2psia)

h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb

n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb

Therefore,

q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb

To calculate the mass flow rate of steam in the cycle, we use the formula

W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s

where 1Kj = 0.947817 Btu

The power output and the rate of heat addition are calculated thus,

W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW

Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s

The thermal efficiency of the cycle can be found thus;

n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343

= 34.3%

5 0
3 years ago
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