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Andre45 [30]
3 years ago
6

After landing on an unexplored Klingon planet, Spock tests for the direction of the magnetic field by firing a beam of electrons

in various directions and by observing the following: Electrons moving upward feel a magnetic force in the NW direction; Electrons moving horizontally North are pushed down; Electrons moving horizontally South-East are pushed upward. He naturally concludes that the magnetic field at this landing site is in which direction?
Physics
1 answer:
BARSIC [14]3 years ago
3 0

Answer:

Magnetic field is in south west direction .

Explanation:

Let us represent various direction by  i , j, k . i representing east , j representing north and k representing vertically upward direction .

magnetic field is represented vectorially as follows

B = B₀ ( - i - j )

In the first case velocity of electron

v = v k

Force = q ( v x B )

= -e [ vk x B₀ ( - i - j ) ]

= evB₀ ( j -i )

Direction of force is north -west .

In the second case velocity of electron

v = vj

Force = -e [ vj x B₀ ( - i - j ) ]

= - evB₀ k

force is downward

In the third case, velocity of electron

v = v( -j +i )

Force = -e [ v( -j +i ) x B₀ ( - i - j ) ]

= 2 evB₀ k

Force is upward.

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A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude
melamori03 [73]
Answers:
(a) B_o  = 0.3466μT
(b) \lambda = 0.4488μm
(c) f = 6.68 * 10^{14}Hz

Explanation:
Given electric field(in y direction) equation:
E_y = 104sin(1.40 * 10^7 x -\omega t)

(a) The amplitude of electric field is E_o = 104. Hence

The amplitude of magnetic field oscillations is B_o =  \frac{E_o}{c}
Where c = speed of light

Therefore,
B_o =  \frac{104}{3*10^8} = 0.3466μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
\frac{2 \pi }{\lambda} = 1.40 * 10^7
\lambda =  \frac{2 \pi}{1.40} * 10^{-7}
\lambda =  0.4488μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = 6.68 * 10^{14}Hz


6 0
3 years ago
Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0
3 years ago
Cellular telephones use what type of wave?
lana [24]
Cellphones use radio frequency waves to transmit sound through the speaker and the microphone.<span> The radio frequency energy that is given off by cell phones is a type of electromagnetic energy.
I hope I helped! =D</span>
4 0
3 years ago
Read 2 more answers
A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t
4vir4ik [10]

The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force  is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is  by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

rms voltage = 78 K v

power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

Learn more about rms current here:-brainly.com/question/20913680

#SPJ4

7 0
2 years ago
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