Signal propagation in the nervous system can be modeled as

a) A resistor network.

b) A wave of electricity that travels down conducting tissue.

c) A series of RC circuits.

Answer:

Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion.

Explanation:

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

λ = 1.8 x 10⁻⁷ m = 180 nm

Answer:

72.16 N

Explanation:

Given:

q₁ = -53.0 μC

q₂ = 105 μC

q₃ = -88.0 μC

q₁ to q₂ = 0.50 m

q₂ to q₃ = 0.95 m

To find:

Net force on q₃

Solution:

First compute net electric field on q₃

E = F/q = k.Q/d²

The formula of electric field at q₃:

E = k.Q / r²

Where    

r is distance

Q is magnitude of charge

k is a constant with a value of 8.99 x 10⁹ N m²/C²

When

q₂ to q₃ = 0.95m and

q₂ = 105 μC then

Find electric field due to charge q₂

E = ( (8.99 x 10⁹)x( 105 x 10⁻⁶) ) / 0.95²

  =  (8990000000  x 0.000105) / 0.9025

  = 943950  / 0.9025

  = 1045927.977839

  = 1.046 x 10⁶ N/C

This interprets that it will act or point away from q₂

q₁ to q₂= 0.50m

q₂ to q₃ = 0.95m and

q₁ = -53 μC then

Find electric field due to charge q₁

E = (8.99 x 10⁹) x (53 x 10⁻⁶) / (0 .50 + 0.95)²

  =  (8990000000  x  0.000053) / (1.45)²

  = 476470 /2.1025

  = 226620.689655

  = 0.227  x 10⁶ N/C

This interprets that it will act or point towards q₁

Since these fields are opposite in direction.

Compute Net Field

Net Field = 1.046 x 10⁶ - 0.227  x 10⁶ N/C

               =  1046000 - 227000

               = 819000

               = 0.819 x 10⁶

               ≈ 0.82 x 10⁶

This interprets that it will act or point away from q₂

Compute force on q3

q₃ E  = 88 x 10⁻⁶ x 0.82 x 10⁶

       = 88000000  x 820000

       = 72160000000000

       = 72.16 N

Force on -ive charge in a field is always in a direction opposite to direction of field

So this interprets that direction of this field will be towards q₂.

Answer:

The safe distance is 199 cm approximately 200 cm

Explanation:

Safe intensity = 1.00 W/m^2

wattage of radar leaked radar = 50.0 W

safe distance from the microwave will be = ?

We know that the intensity of a wave radiated uniformly in all direction is given as

[tex]I[/tex] = [tex]\frac{W}{A}[/tex]

where

W is the wattage of the leaked radar

A is the radial area, which is the area of a sphere that encapsulates the region through which this wave spreads uniformly.

From the equation above,

[tex]A[/tex] = [tex]\frac{W}{I}[/tex] = 50/1 = 50 m^2

But the area of this sphere [tex]A[/tex] = [tex]4\pi r^{2}[/tex]

where

r is the safe distance from the radar source

substituting for the area, we have

50 = 4 x 3.142 x [tex]r^{2}[/tex]

50 = 12.568 [tex]r^{2}[/tex]

[tex]r^{2}[/tex] = 50/12.568 = 3.978

r = [tex]\sqrt{3.978}[/tex] = 1.99 m = 199 cm ≅ 200 cm

Answer:

Explanation:

the resistance of a electrical device

R = V² / P where V is volt and P is power .

The devices are in parallel so same volt will apply on them

So R₁ = 12² / 32.5 = 4.431 ohm

R₂ = 12² / 2 x 10³ = .072 ohm

when they are in series

Common Current in them = 12 / 4.431 + .072

= 2.6649 A

power consumed by first device when they are in series

= current² x resistance

= 2.6649² x 4.431 = 31.46 W

power consumed by other

= 2.6649² x .072  = .511 W

Answer:

The potential difference at the customer's house is 117.1 V.

Explanation:

a) The potential difference at the customer's house can be calculated as follows:

[tex] \Delta V_{h} = \Delta V_{p} - \Delta V_{l} [/tex]

Where:

[tex]V_{h}[/tex]: is the potential difference at the customer's house

[tex]V_{p}[/tex]: is the potential difference from the main power lines = 120 V

[tex]V_{l}[/tex]: is the potential difference from the lines

[tex] \Delta V_{h} = \Delta V_{p} - IR [/tex]

The resistance, R, is:

[tex]\frac{0.109 \Omega}{300 m}*2*34.0 m = 0.025 \Omega[/tex]

Now, the potential difference at the customer's house is:

[tex]\Delta V_{h} = 120 V - 116A*0.025 \Omega = 117.1 V[/tex]

Therefore, the potential difference at the customer's house is 117.1 V.

I hope it helps you!

Explanation:

Ohm's law gives the relationship between current, voltage, and resistance. Its mathematical form is given by :

V = IR

I is current and R is resistance

Resistance resists the flow of electric current in a circuit. When the amount of applied voltage is increased, it will not change the resistance. It increases the current. Hence, the correct option is (C).

Answer:

3.94

Explanation:

divide total mass by the number of blocks since they are identical

Answer:

3.94

Explanation:

You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks

Answer:

It emits alpha and beta particles.

Explanation:

EDG 20

Answer:

the answer is b, calcium-45

Explanation:

Answer:

Explanation:

magnetic field B = (3 i + 8 x 2 j )x 10⁻³ T

= (3 i + 16 j )x 10⁻³ T

L = - i ( unit length of conductor )

Force F = I ( L x B ) , I is current

= 5 [ - i x ( 3i + 16 j ) 10⁻³]

= 5 ( - 16 k ) x 10⁻³

F = - 80 x 10⁻³ k

Answer:

[tex]E=6.01\times 10^{-3}\ J[/tex]

Explanation:

It is given that,

Side of a square loop, l = 24 cm = 0.24 m

Resistance of loop, R = 6.1 ohms

Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms

We need to find the electrical energy dissipated in this process.

Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :

[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]

If I is induced current then,

[tex]V=IR[/tex]

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]

Power is given by voltage times current. So,

[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]

Now, energy is given by the product of power and time. So,

[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]

Now putting all the values in above formula. So,

[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]

So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.

The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".

According to the question,

Square loop's side, l = 24 cm or,

                                  = 0.24 m

Loop's resistance, R = 6.1 ohms

Initial magnetic field = 0.665 T

Final magnetic field = 0

We know the relation,

→ V = - [tex]\frac{dB}{dt}[/tex]

      = [tex]\frac{BA}{t}[/tex]

Also we know,

Current, V = IR

               I = [tex]\frac{V}{R}[/tex]

                 = [tex]\frac{BA}{tR}[/tex]

Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,

                          = [tex]\frac{(BA)^2}{tR}[/tex]

By substituting the values,

                          = [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]

                          = 6.01 × 10⁻³ J

Thus the response above is correct.                                  

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Answer:

Options A, B as well as C are the correct choices.

Explanation:

This form of learning isn't always linked to the perhaps one given choice. And the response to the above will still be the appropriate one.

Answer:

Explanation:

Given,

Mass ( m ) = 70 kg

Distance ( d ) = 4.5 m

Time taken ( t ) = 4 seconds

Power = ?

Now, applying the formula to find power:

[tex]power = \frac{work \: done}{time \: taken} [/tex]

[tex] = \: \frac{f \: \times \: d}{t} [/tex]

[tex] = \frac{m \: \times \: g \: \times \: d}{t} [/tex]

Plugging the values:

[tex] = \frac{70 \times 9.8 \times 4.5}{4} [/tex]

Calculate the products

[tex] = \frac{3087}{4} [/tex]

Divide:

[tex] = 771.75 \: watt[/tex]

Hope this helps...

Best regards!!

Answer:

Ok, sabemos que la velocidad inicial de la pelota es 15m/s.

Desconocemos la posición inicial a la que es lanzada la pelota, pero vamos a suponer que es a una altura igual a cero, es decir, la pelota es lanzada al ras del suelo.

Una vez lanzada, la única fuerza actuando en la pelota es la gravitatoria, entonces la aceleración de la pelota es:

a = -g = -9.8m/s^2

El signo negativo es por que esta aceleración apunta hacia abajo.

Ahora, para la velocidad, necesitamos integrar sobre el tiempo.

v(t) = (-9.8m/s^2)*t + v0

donde v0 = 15m/s

v(t) = (-9.8m/s^2)*t + 15m/s.

De aca podemos obtener el tiempo en el que la pelota llega a la altura máxima, que es el punto donde la velocidad es igual a cero.

0 = (-9.8m/s^2)*t + 15m/s.

t = (15/9.8)s = 1.53 s

Ahora, para la ecuación de la posición integramos la ecuación de la velocidad sobre el tiempo:

p(t) = (1/2)(-9.8m/s^2)*t^2 + 15m/s*t + p0

donde p0 es la pocision inicial, pero arriba dijimos que era igual a cero, entonces la ecuación queda:

p(t) = (-4.5m/s^2)*t^2 + 15m/s*t

ahora reemplazamos t por el tiempo que encontramos antes, y descubrimos que:

p(1.53s) =  (-4.5m/s^2)*(1.53s)^2 + 15m/s*1.53s = 12.41m

La máxima altura que alcanza la pelota es 12.41 metros arriba del punto desde el que se la lanzo.

Ahora, el tiempo total que esta en el aire puede ser calculado de tal forma que la posición vuelva a ser cero, es decir, la pelota llega a la misma altura desde la que fue lanzada inicialmente (y es agarrada por la persona, podemos suponer)

Entonces:

p(t) = 0 =  (-4.5m/s^2)*t^2 + 15m/s*t

Ahora resolvemos la eq cuadrática, usando la eq. de Bhaskara:

[tex]t = \frac{-15 +- \sqrt{15^2 - 4*(-4.5)*0} }{-2*4.5} = \frac{-15 +-15}{-9.8}[/tex]

Entonces las soluciones son:

t = (-15 + 15)/-9.8 = 0s

t = (-15 - 15)/-9.8 = 3.06s

Tomamos la segunda solución, ya que la primera corresponde al tiempo inicial.

Entonces concluimos con que la pelota estuvo 3.06 segundos en el aire.

Answer:

The magnetic field remains the same.

Explanation:

If a solenoid has length L, N windings, and radius b, then the magnetic field inside the solenoid is given by :

[tex]B=\mu_o NI[/tex]

[tex]\mu_o[/tex] is magnetic permeability

If the radius of the solenoid were doubled and all other quantities remained the same, the magnetic field inside the solenoid would remains the same as it is independent of the radius of the solenoid.

Answer:

5.5 m/s^2

Explanation:

I believe this is the answer > using the formula a= v-v0/t

Hope this helps!

Answer:

-1 m/s²

Explanation:

Given:

Δx = 32 m

v₀ = 10 m/s

t = 4 s

Find: a

Δx = v₀ t + ½ at²

32 m = (10 m/s) (4 s) + ½ a (4 s)²

a = -1 m/s²

Answer:

5.23 C

Explanation:

The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.

Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁

B₁ = 0.670 T and B₂ = 0 T

ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T

A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m

A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²

ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt

Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire =  1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m

R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω

So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω  

IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C

Since ΔQ = It = 5.232 C ≅ 5.23 C

So the charge is 5.23 C

The induced emf through wire depends on the current flow (indirectly with charge flow as well).

The value of charge moving past a point in the coil during its operations is 5.23 C.

The region in a space where a particle experiences some magnetic force, is known as the magnetic field.

Given data -

The diameter of the circular loop is, d = 13.2 cm = 0.132 m.

The change in magnetic field strength is, ΔB = 0.670 T.

The new diameter of the wire is, d' = 2.25 mm = 2.25 × 10³ m.

The resistivity of the wire is, [tex]\rho = 1.68 \times 10^{-8} \;\rm \Omega.m[/tex].

The current in the wire is given by the following expression,

I = ε/R

Here,

ε is the induced emf in the wire.

R is the resistance of the wire.

And the expression for the induced emf of the wire is given as,

ε = -ΔΦ/Δt

Here,

ΔΦ is the change in magnetic flux. And its expression is,

ε = A × ΔB

Here,

A is the area of the loop. And its value is, A = πd²/4.

Solving as,

A = π(0.132 m)²/4

A = 0.01368 m²
A = `1.368 × 10⁻² m²

Now, calculating the induced emf as,

ε = ΔΦ/Δt

ε = A × ΔB/Δt

ε = 1.368 × 10⁻² m² × (0.670 T)/Δt

ε = 0.9166 × 10⁻² Tm²/Δt

Now, the resistance R of the circular wire is,

R = ρ × L/A'

Here,

L is the length of the wire and its value is. L = πd .

And A' is the new cross-sectional area of wire,

A' = πd'²/4

So, the resistance is,

R = ρ × L/A'

R = 1.68 x 10⁻⁸  ×( π × 0.132 m) ÷ π(2.25 × 10⁻³ m)²/4 =

R = 0.88704/5.0625

R = 0.1752 × 10⁻² Ω

R = 1.752 × 10⁻³ Ω

Now, the current flow (I) in the wire is given as,

I = ε/R

I = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω  

And obtaining the value of charge from the expression of current as,

Q = IΔt
Q = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω

Q = 0.5232 × 10 C

Q = 5.23 C

Thus, we can conclude that the value of charge moving past a point in the coil during its operations is 5.23 C.

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Answer:

-4m/s

Explanation:

use the formula

[tex]f = ma[/tex]

where f-force

m-mass

a-accleration

so

1kN=1000N

so apply

4000=1000×a

a=4m/s

(the negative is because the car was braking)

Answer:

Your answer is -4 m/s^2

Explanation:

Set Up: Let +x be the direction the car is traveling.  

List the known & unknown quantities:  

   m = mass of the car = 1000 kg

  υ = 30 m/s

  Fx = –4 kN = –4000 N (negative since it is a braking force)

  ax = acceleration =?  

Solve: Use Newton’s second law of motion.  

Fx=max    

ax=Fx/m = −4000 N /1000 kg = −4000 kg·m/s^2 / 1000 kg =−4m/s^2

a).  frequency = (speed) / (wavelength)

The speed of light is around 3 x 10⁸ m/s.

For 380 nm (violet light), frequency = (3 x 10⁸ m/s) / (380 x 10⁻⁹ m)

Frequency = 7.89 x 10¹⁴ Hz

For 760 nm (red light), frequency = (3 x 10⁸ m/s) / (760 x 10⁻⁹/s)

Frequency = 3.94 x 10¹⁴ Hz

The ratio is 2 .

That's 1 octave, or 0.3 of a decade.

b).  The ratio of highest/lowest sounds is (20,000 Hz/20 Hz) = 1,000

That's 3 decades, or about 10 octaves.

===> Speaking logarithmically ( ! ), ears are sensitive to a range of sound frequencies that's 10 times as wide as the range of light frequencies that eyes can detect.

Answer:

Multiply the x values with -1.

Explanation:

By multiplying the numbers by one, you are changing them to be the opposite of their original state.

You multiply the numbers that are in the x-value column because you are reflecting the image over the y-axis.

Hope this helped and good luck!

Answer:

Multiply each x-value in the table by -1

Explanation:

Answer:

8996kg*m/s/s

Explanation:

Given:

a=26m/s/s

m=346kg

Required:

f=?

Formula:

f=m*a

Solution:

f=346kg*26m/s/s

f=8996kg*m/s/s

Hope this helps ;) ❤❤❤

Answer:

[tex]\boxed{F = 8996 \ Newton}[/tex]

Explanation:

Given:

Mass = m = 346 kg

Acceleration = a = 26 m/s²

Required:

Force = F = ?

Formula:

F = ma

Solution:

F = 346 * 26

F = 8996 Newton

Answer:

a = 30 km / h²

Explanation:

Dado que

Velocidad inicial, u = 70 km / h

Velocidad final, v = 95 km / h

Tiempo, t = 30 s = 0.1 h

Lo sabemos

v = u + a t

a = aceleración

Ahora poniendo los valores en la ecuación anterior

[tex]95 = 70 + a \ times 0.1 [/tex]

[tex]a = \ dfrac {95-70} {0.1} = 30 \ km / h ^ 2 [/tex]

Por lo tanto, la aceleración será

a = 30 km / h²

Answer:

personally I'd say C by do not know if that is the exact answer

Answer:

Net force = 0

Explanation:

Short answer: if the middle book is not acceleration in any direction, the net force is zero.

Long answer: refer to the attached free-body diagram (FBD) to see why the net force is zero.  Skills to draw FBD are  essential in solving problems in statics.

Top book exerts -mg (downwards) on the middle book.

Middle book exerts -mg (downwards) on the bottom book.

Total downward force is -mg-mg = -2mg

By Newton's third law, when an object is not in motion, reaction equals action of -2mg in the opposite direction, therefore the reaction (upwards) is +2mg.

This makes the net force of -2mg (downwards) + 2mg (upwards) =0

On a horizontal table, three books are resting and balanced. The net force on the middle book will be equal to zero.

Equilibrium is the condition of an object when two or more opposing forces, whether internal, external or a combination of both, act on the body and cancel one another out to maintain the object in the same state as it was. The Latin term for weight or balance, libra, serves as the origin of the word equilibrium.

According to the question :

The net force is 0 if the middle book does not accelerate in any direction.

The Middle book is impacted negatively (downwards) by the top book.

The middle book pulls down on the lower book by -mg.

The total downward force is -2mg = (-mg-mg)

According to Newton's third rule, a reaction when an item is not moving equals an action of -2mg in the opposite direction, hence the reaction (upwards) is a response of +2mg.

Consequently, the net force of -2mg (downwards) + 2mg (upwards) = 0 is created.

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Answer:

Explanation:

In physics, there are two types of physical quantities namely the fundamental and the derived quantities. Fundamental quantities are independent quantities on which derived quantities depends on. Length, mass and time are examples of fundamental quantities.

The SI unit of length is meters. A meter is a multiple unit. Its submultiple units are centimetres (10⁻²metres), kilometres (10³metres), decimetres (10⁻¹metres) etc

The SI unit of mass is kilogram (kg). The only sub multiple unit used in real-life situation is grams.

1 kg = 100 grams

The SI unit of time is seconds. The multiple units are the minutes, hours, weeks, days and years.

1 minute = 60 seconds

1 hour = 3600 seconds

1 day = 24 * 3600 = 86,400 secs

Explanation:

let us use the explanation below to get the intuition so desired;

According to Faraday's law of electro magnetic induction, when ever a coil/conductor is made to rotate in a magnetic field, voltage or emf is created and current is produced, in the long run energy has be produced or converted.

The conversion of this energy is made possible by the motion of the coil/conductor is the magnetic field, just by the  motion of the conductor cutting through the magnetic field, thus creating electro motive force(E.M.F) hence producing current, and ultimately energy is created

Answer with explanation:

Complete question is provided in the attachment below.

There are 3 undefined term in geometry :1) A Point  2)  A line 3) A Plane.

When two lines are parallel they never meet , so the requirement to define them is lines and a plane on which they lie.

While when two lines are perpendicular , they intersect each other at a point by making a right angle between them.

So it required lines and a point to define it.

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]

We need to find the thrust and the force. The mass of the liquid displaced is given by :

[tex]m=dV[/tex]

V is volume

Weight of the displaced liquid

W = mg

[tex]W=dVg[/tex]

So,

[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]

So, the thrust and the force is 1562.53 N.