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3 years ago

How often do the earth's magnetic poles switch?

1 answer:

3 0

Reversals are the rule, not the exception. Earth has settled in the last 20 million years into a pattern of a pole<span> reversal about every 200,000 to 300,000 years, although it has been more than twice that </span>long<span> since the last reversal.</span>

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

2 years ago

1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.

marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

$B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB$

<h3>Speed of sound at the given temperature</h3>

$v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s$

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

$f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )$

where;

-v₀ is velocity of the observer moving away from the source
-vs is the velocity of the source moving towards the observer
fs is the source frequency
fo is the observed frequency
v is speed of sound

$f_0 = f_s(\frac{v-v_0}{v- v_s} )$

$f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz$

Learn more about intensity of sound here: brainly.com/question/17062836

3 0

1 year ago

Superman is flying 54.5 m/s when he sees

Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations $d=v_{0} t+\frac{at^{2} }{2}$ and $v_{f} =v_{0} +at$ where $d$ is distance, $v_{0}$ is the initial velocity, $v_{f}$ is the final velocity, $t$ is time, and $a$ is aceleration.

Superman's initial velocity is $v_{0}=54.5\frac{m}{s}$ , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know $d$ , $v_{0}$ and $t$ , we have to find the aceleration $a$ in order to find $v_{f}$ .

From the equation $d=v_{0} t+\frac{at^{2} }{2}$ we have to clear $a$ , getting the equation as follows: $a=\frac{2(d-v_{0}t) }{t^{2} }$ .

Substituting the values:

$a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}$

To find $v_{f}$ we use the equation $v_{f} =v_{0} +at$ .

Substituting the values:

$v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}$

5 0

3 years ago

for a certain chemical reaction, the reactants contain 52 kj of potential energy and the products contain 32 kj how much energy

Luden [163]

B, is the answer. (20 kj is released)

7 0

2 years ago

Read 2 more answers

In the bohr model, what is the ratio of its kinetic energy to its potential energy?

stich3 [128]

The centrifugal force C = mv^2/r = kq^2/r^2 = P centripetal force. m is the electron mass, q are the proton and electron charges (opposites), and r is the Bohr radius.

Thus 1/2 mv^2/r = 1/2 kq^2/r^2 and KE = 1/2 mv^2 = 1/2 kq^2/r = 1/2 PE

Therefore KE/PE = 1/2, no matter what state the electron is in.

8 0

2 years ago