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BaLLatris [955]
2 years ago
4

A beam of electrons is accelerated through a potential difference of 10 kV before entering a region having uniform electric and

magnetic fields that are perpendicular to each other and perpendicular to the direction in which the electron is moving. If the magnetic field in this region has a value of 0.010 T, what magnitude of the electric field is required if the particles are to be undeflected as they pass through the region?
Physics
1 answer:
Hunter-Best [27]2 years ago
8 0

Answer:

The magnitude of the electric field is 592.67 V/m

Explanation:

The kinetic energy of an electron moving with speed v is:

K.E. = (1/2)mv²

Also,

The kinetic energy for an electron in a electrical field:

K.E. = eV

Where e is the charge of the electron and V is the potential difference

Thus,

<u>(1/2)mv² = eV </u>

So,

v=\sqrt{\frac {2eV}{m}}

Mass of electron = 9.11*10⁻³¹ kg

Charge on electron = 1.6*10⁻¹⁹ C

Given: voltage = 10 kV = 10000 V

So,

v=\sqrt{\frac {2\times (1.6\times 10^{-19})\times 10000}{9.11\times 10^{-19}}}

<u>v = 5.9267*10⁷ m/s</u>

The two fields are the crossed fields, So,

E = v×B

Given B = 0.010 T

E = 5.9267*10⁷×0.010 V/m

E = 5.9267*10⁵ V/m

Or,

<u>E = 592.67 kV/m</u>

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An object executes simple harmonic motion with an amplitude A. (Use any variable or symbol stated above as necessary.) (a) At wh
valentina_108 [34]

Answer:

(a) x=ASin(ωt+Ф₀)=±(√3)A/2

(b) x=±(√2)A/2

Explanation:

For part (a)

V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)

Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3

x=ASin(ωt+Ф₀)=±(√3)A/2

For part(b)

U=0.5E and U+K=E→K=0.5E

E=K(Max)

(1/2)mv²=(0.5)(1/2)m(Vmax)²

V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4

x=±(√2)A/2

7 0
2 years ago
According to Newton's Law of Universal Gravitation, which of the following would cause the attractive force between a planet and
vladimir1956 [14]

Answer:

it is either a or c

Explanation:

4 0
2 years ago
As the driver steps on the gas pedal, a car of mass 1 140 kg accelerates from rest. During the first few seconds of motion, the
krok68 [10]

Answer:

(a) KE=16405.215 J

(b) P = 6309.6981 W

(c) Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

Explanation:

Given:

mass of car, m = 1140 kg

expression of acceleration, a=1.14t-0.210t^2+0.240t^3

where "t" is time in seconds

initial time, t_i=0 s

final time, t_f=2.6 s

(a)

We know,

\frac{dv}{dt} =a

dv=a.dt

v=\int\limits^{2.6}_0 {1.14t-0.210t^2+0.240t^3} \, dt

v=5.3648 m.s^{-1}

Kinetic Energy

∴KE= \frac{1}{2} m.v^2

KE=\frac{1}{2}\times 1140\times 5.3648^2

KE=16405.215 J

(b)

We know,

Power

P= \frac{\Delta KE}{\Delta t}

P=\frac{16405.215}{2.6}

P = 6309.6981 W

(c)

Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

3 0
2 years ago
A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0
3 years ago
A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor
Alex17521 [72]

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0
2 years ago
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