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xxTIMURxx [149]
3 years ago
6

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.5 meters

in the first 2.84 μs after it is released. What is magnitude of the electric field? What is the direction of the magnetic field? Are we justified if ignoring gravity on the electron in this situation?
Physics
1 answer:
dem82 [27]3 years ago
8 0

Answer:

The electric and magnetic field are 6.34 N/C and 2.11\times10^{-8}\ T.

Explanation:

Given that,

Distance = 4.5 m

Time = 2.84 μs

We need to calculate the acceleration

Using equation of motion

The distance covered by the electron is

s=ut+\dfrac{1}{2}at^2

When, the electron at rest

s = \dfrac{1}{2}at^2

Where, s = distance

a = acceleration

t = time

Put the value into the formula

4.5=\dfrac{1}{2}\times a\times(2.84\times10^{-6})^2

a=\dfrac{2\times4.5}{(2.84\times10^{-6})^2}

a=1.116\times10^{12}\ m/s^2

We need to calculate the electric field

Using formula of the electric field

E=\dfrac{F}{q}

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{9.1\times10^{-31}\times1.116\times10^{12}}{1.6\times10^{-19}}

E=6.34\ N/C

We need to calculate the magnetic field

Using formula of magnetic field

B = \dfrac{E}{c}

Put the value into the formula

B=\dfrac{6.34}{3\times10^{8}}

B=2.11\times10^{-8}\ T

According to the right hand rule,

The direction of magnetic field is outward because the direction of force is upward.

Hence, The electric and magnetic field are 6.34 N/C and 2.11\times10^{-8}\ T.

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Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman
Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0
3 years ago
Types of cell combination? in electricity​
Grace [21]

Answer:

Series and Parallel

7 0
3 years ago
Which of the following motions has a straight line?
skad [1K]
The answer would probably be B.
8 0
3 years ago
What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g
aalyn [17]

Answer:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle \theta, given by:

f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)

Where:

Z_1 e represent the charge of the projectile (Z1=2)

Z_2 e is the target charge (z2=79)

K= 5x10^6 eV represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

t= 10^{-8] m represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

n =\frac{\rho N_A N_M}{M_g}

Where:

\rho = 19.3 g/m^3= 19300 Kg/m^3

N_A = 6.022 x10^{23} molecules/mol the Avogadro's number

N_M = 1 represent the atoms per molecule

M_g = 197 g/mol = 0.197 Kg/mol represent the molecular weigth

If we replace we got:

n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness t=10^{-8}m

And using the first formula we got:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

4 0
3 years ago
What is the net force required to give an automobile of mass 1600 kg an acceleration of 4.5 m/s2 ?
Svetach [21]
Net force is calculated using the equation

F=ma

where m is mass and a is acceleration.

So, F=(1600kg)(4.5m/s^2)
= 7200N
4 0
3 years ago
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