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Allushta [10]
3 years ago
15

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

n electron with kinetic energy of 1 keV? Compare with the gravitational force on the electron.
Physics
1 answer:
vampirchik [111]3 years ago
4 0

Answer:

F = 1.5 \times 10^{-16} N

this force is 1.68 \times 10^{13} times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

KE = 1 keV

KE = 1 \times 10^3 (1.6 \times 10^{-19}) J

KE = 1.6 \times 10^{-16} J

now the speed of electron is given as

KE = \frac{1}{2}mv^2

now we have

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

v = 1.87 \times 10^7 m/s

now the maximum force due to magnetic field is given as

F = qvB

F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})

F = 1.5 \times 10^{-16} N

Now if this force is compared by the gravitational force on the electron then it is

\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}

\frac{F}{F_g} = 1.68 \times 10^{13}

so this force is 1.68 \times 10^{13} times more than the gravitational force

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A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
My name is Ann [436]

Answer:

a)   C = 4,012 10⁻¹⁴ F, b)  Q = 1.6 10⁻¹¹ C , c)   U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

       C = k e₀ A / d

Let's calculate

       C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

       C = 4,012 10⁻¹⁴ F

b) let's look  the charge

        C = Q / ΔV

         Q = C ΔV

         Q = 4,012 10⁻¹⁴ 400

         Q = 1.6 10⁻¹¹ C

c) The stored energy

        U = ½ C ΔV²

        U = ½ 4,012 10⁻¹⁴  400²

        U = 3.21 10⁻¹¹ J

4 0
3 years ago
A ball rolls off a desk at a speed of 3 m/s and lands .40 seconds later. How far from the base of the desk does the ball land?
Salsk061 [2.6K]

Is the velocity constant? Is there any friction?

3 meters per second

then after 40 seconds it must 3*40 = 120 meters

120 meters or 0.12 km if you will

7 0
3 years ago
What experimental evidence led to the development of this atomic model from the one before it?
Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0
3 years ago
Read 2 more answers
A wall (of thermal conductivity 0.98 W/m · ◦ C) of a building has dimensions of 3.7 m by 15 m. The average inside and outside
disa [49]

Answer:

Answer:

the amount of energy flowing is 1.008x10⁹J

Explanation:

To calculate how much heat flows, the expression is the following:

Where

K=thermal conductivity=0.81W/m°C

A=area=6.2*12=74.4m²

ΔT=30-8=22°C

L=thickness=8cm=0.08m

t=time=16.9h=60840s

Replacing:

Explanation:

4 0
2 years ago
A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
vladimir1956 [14]

Answer:

Force, F=2.27\times 10^4\ N

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

F.d=\dfrac{1}{2}m(v^2-u^2)

Finally, it stops, v = 0

F.d=-\dfrac{1}{2}m(u^2)

F=\dfrac{-mu^2}{2d}

F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}

F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

7 0
3 years ago
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