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balandron
4 days ago
13

Through how many volts of potential difference must an electron, initially at rest, be accelerated to achieve a wave length of 0

.27nm? A) 4.6 V
B) 0.046 V
C) 0.27 V
D) 4.6 MV
E)4.6 kV
Physics
1 answer:
kicyunya [2.9K]4 days ago
4 0
20.7 volts. The mass of an electron is 9.1 x 10⁻³¹ kg, and its wavelength is 0.27 x 10⁻⁹ m. The velocity of the electron can be determined using de Broglie's equation λ mv = h. Substituting the known values, we arrive at v = 2.7 x 10⁶ m/s. The potential difference through which the electron accelerates is noted, with the charge on an electron being 1.6 x 10⁻¹⁹ C. According to the conservation of energy, (0.5) mv² = q ΔV leads to ΔV = 20.7 volts.
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An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT. The electron follows a helical p
Softa [2661]

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v  = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = \frac{2\pi m}{qB} = \frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})} = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

6 0
25 days ago
A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
kicyunya [2911]

Answer:

W = -510.98 J

Explanation:

Force = 43 N, 61° SW

Displacement = 12 m, 22° NE

The work done is calculated using:

W = F*d*cos(A)

where A is the angle between the applied force and displacement.

The angle A between the force and displacement is determined as A = 61 + 90 + 22 = 172°

Hence, W = 43 * 12 * cos(172)

This results in W = -510.98 J

The negative result indicates that the work is done contrary to the direction of the force applied.

6 0
28 days ago
A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.
serg [3228]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where

k represents Coulomb's constant

q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C denotes the charge magnitude on the first cork

q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) 2.69\cdot 10^{13}

The total charge of the negative cork is

q_2 = -4.3 \cdot 10^{-6}C

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

e=-1.6\cdot 10^{-19}C

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

q_2 = Ne

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}

D) 3.75\cdot 10^{13}

The overall charge on the positive cork is

q_1 = +6.0\cdot 10^{-6}C

<pthe charge="" of="" a="" single="" electron="" is="" known="" to="" be="">

e=-1.6\cdot 10^{-19}C

<pthe total="" charge="" of="" the="" positive="" cork="" results="" from="" n="" excess="" electrons="" which="" can="" be="" depicted="" as="">

q_1 = -Ne

<pby calculating="" for="" n="" we="" derive="" the="" number="" of="" electrons="" cork="" has="" lost:="">

N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}

</pby></pthe></pthe></pafter></pthe></pwe></pthus></pin></pmore></pas>
6 0
27 days ago
Read each scenario below. Then select the answer that best completes each sentence.
Ostrovityanka [2814]

Answer:

The power used by raul's microwave must match the power consumed by katrina's because both microwaves took different durations to accomplish the same heating task.

Explanation:

The power output from a car engine is equivalent to that of a bicycle since both perform the same amount of work over time. Both raul and katrina shared a frozen meal, heating each portion in different microwaves. Katrina's portion was warm in one minute, whereas raul's portion required two minutes. Therefore, the power utilized by raul's microwave aligns with that of katrina's, given that it took longer to achieve the same result.

7 0
1 month ago
Read 2 more answers
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Softa [2661]
Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
The velocity at the stagnation point is zero.

The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Thus, the answer is 2.2 psi

5 0
12 days ago
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