g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be

Answer:

1.3515x10^5pa

Explanation:

Plss see attached file

Answer:

1.24m

Explanation:

See attached file

Answer:

The energy density is  [tex]Z = 5.4 2 *10^{-5 } \ J/m^3[/tex]

Explanation:

From the question we are told that

    The electric vector is  [tex]E = 3500 \ V/m[/tex]

Generally the energy vector is mathematically represented as

      [tex]Z = 0.5 * \epsilon_o * E^2[/tex]

Where  [tex]\epsilon_o[/tex] is the permitivity of free space with the value  [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

      [tex]Z = 0.5 * 8.85 *10^{-12} * 3500^2[/tex]

      [tex]Z = 5.4 2 *10^{-5 } \ J/m^3[/tex]

Answer:

Explanation:

The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where;

R is the resistance of the material

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the area = πr² with r being the radius

[tex]R = \rho L/\pi r^{2}[/tex]

If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L

The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex]

since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become

[tex]R_1 = \frac{\rho(2L)}{A}[/tex]

[tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex]

Taking the ratio of both resistances, we will have;

[tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex]

This shoes that the new resistance of the wire will be twice that of the original wire

Answer:

electric field in the wide wire is

E₂ =[tex]\frac{E}{4}[/tex]

Explanation:

given

length of the copper wire = L

radius of the copper wire r₁ = b

length of the second copper wire = L

radius of the second copper wire r₂ = 2b

electric field in the narrow wire = E₁=E

recall

resistance R = ρL/A

where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.

Resistance of narrow wire, R₁

R₁ = ρL/A

where A  = πb²

R₁ = ρL/πb²---------- eqn 1

Resistance of wide wire, R₂

R₂ = ρL/A

where A = π(2b)²

R₂ = ρL/π(2b)²

R₂ = ρL/4πb²-------------- eqn 2

R₂ = ¹/₄(ρL/πb²)

comparing eqn 1 and 2

R₁ = 4R₂

calculating the current in the wire,

I = E/(R₁ + R₂)

recall

R₁ = 4R₂

∴ I = E/(4R₂ + R₂)

I = E/5R₂

calculating the potential difference across R₁ & R₂

V₁ = IR₁

I = E/5R₂

∴ V₁ = ER₁/5R₂

R₁ = 4R₂

V₁ = 4ER₂/5R₂

∴V₁  = ⁴/₅E

potential difference for R₂

V₂= IR₂

I = E/5R₂

∴ V₂ = ER₂/5R₂

V₂ = ER₂/5R₂

∴V₂  = ¹/₅E

so, electric field E = V/L

for narrow wire E₁ = V₁/L ----------- eqn 3

for wide wire, E₂ = V₂/L------------ eqn 4

compare eqn 3 and 4

E₂/E₁ = V₂/V₁( L is constant)

E₂/E₁ = ¹/₅E/⁴/₅E

E₂ = E₁/4

note E₁ = E

∴E₂ =[tex]\frac{E}{4}[/tex]

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s

final velocity of bullet [tex]v_{2}[/tex] = 140 m/s

loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]

==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = [tex]\frac{1}{2} mv^{2}[/tex]

70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]

566.28 = [tex]v^{2}[/tex]

[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

Explanation:

F = kx

F = (1500 N/m) (0.30 m)

F = 450 N

Answer:

I think the answer probably be B

What explains why a prism separates white light into a light spectrum ?

C. The different colors that make up a white light have different refractive indexes in glass.

✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).

✔ That's why purple is more deflected so is lower than red radiation.  

Answer:

230N/m

Explanation:

Pls see attached file

Answer:

a) 17.05 mph

b) 54.7°  northeast direction

c) 10.71 mph

The direction is -22.58° relative to the east.

To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

[tex]V_{y}[/tex] = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

[tex]V_{x}[/tex] = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = [tex]\sqrt{V^{2} _{y}+V^{2} _{x} }[/tex]

==> [tex]\sqrt{13.89^{2} +9.89^{2} }[/tex] = 17.05 mph  This is your airspeed

b) To get your direction, we use

tan ∅ = [tex]V_{y}[/tex] /[tex]V_{x}[/tex]

tan ∅ = 13.89/9.89 = 1.413

∅ = [tex]tan^{-1}[/tex](1.413) = 54.7°  northeast direction

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

[tex]V_{y}[/tex] = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = [tex]\sqrt{V^{2} _{y}+V^{2} _{x} }[/tex]

==> [tex]\sqrt{4.11^{2} +9.89^{2} }[/tex] = 10.71 mph  This is your airspeed

Your direction will be,

tan ∅ = [tex]V_{y}[/tex] /[tex]V_{x}[/tex]

tan ∅ = -4.11/9.89 = -0.416

∅ = [tex]tan^{-1}[/tex](-0.416) = -22.58°  this is the angle you'll travel relative to the east.

To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.

Answer:

Explanation:

 This problem bothers on  application of the expression for motion emf which is expressed as

[tex]E= Blv[/tex]

where B= magnetic field in Tesla

           l= length of the conductor

           v= speed of conductor

Given data

l= 2 meters

v= 6 m/s

B= 5 Tesla

Applying the formula we have

[tex]E=5*2*6= 60mV[/tex]

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

The drift velocity of the electrons should be 0.44 mm/s.

Since the diameter is 0.081 in

So, the  radius = r = 0.081 inch/2

= 0.0405 inch

Now the conversion of inches to cm should be

= 0.0405*2.54

= 0.10287 cm

Now

area = π r2

= 3.14 * (0.10287)2

= 0.0332

Now the velocity should be

v = i/(nqA)

= 20/(8.5e22*1.60e-19*0.0332)

= 0.044 cm/s

= 0.44 mm/s

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Answer:

The  kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]

Explanation:

From the question we are told that

       The  radius of the orbit is  [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]

       The gravitational force is  [tex]F_g = 6600 \ N[/tex]

The kinetic energy of the satellite is mathematically represented as

       [tex]KE = \frac{1}{2} * mv^2[/tex]

where v is the speed of the satellite which is mathematically represented as

     [tex]v = \sqrt{\frac{G M}{r^2} }[/tex]

=>  [tex]v^2 = \frac{GM }{r}[/tex]

substituting this into the equation

      [tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]

Now the gravitational force of the planet is mathematically represented as

      [tex]F_g = \frac{GMm}{r^2}[/tex]

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     [tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]

=>    [tex]KE = \frac{ 1}{2} *F_g * r[/tex]

substituting values

       [tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]

         [tex]KE = 7.59 *10^{10} \ J[/tex]

B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.

Answer:

3.21×10⁻³ Ωm

Explanation:

Applying,

R = Lρ/A................... Equation 1

Where R = Resistance of the wire, L = Length of the wire, ρ = Resistivity of the wire, A = cross sectional area of the wire.

Make ρ the subject of the equation

ρ = RA/L................... Equation 2

Given: R = 0.31 Ohms, A = 0.003 m², L = 0.29 m

Substitute into equation 2.

ρ = 0.31(0.003)/0.29

ρ  = 3.21×10⁻³ Ωm

Answer:

amplitudes

Explanation:

In everyday physics we define the amplitude of a wave as the maximum (this can also be called the highest)displacement or distance moved by a point on a given vibrating body or wave as measured from its equilibrium position. The key idea in defining the amplitude of a wave motion is the idea of a 'maximum displacement from the position of equilibrium'.

Given the equations;

E= EoSin(kx - ωt)y

B= Bosin(kx- ωt)z

Both Eo and Bo refer to the maximum displacement of the electric and magnetic field components of the electromagnetic wave. This maximum displacement is known as the amplitude of the electric and magnetic components of the electromagnetic wave.

Answer:

Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.

Explanation:

The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.  

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

Answer:

1.5x10^-1 V

Explanation:

See attached file

Answer:

The magnitude of the induced emf in the coil is 15.3 mV

Explanation:

Given;

number of turns, N = 20 turns

Area of each coil, A = 0.0015 m²

initial magnitude of magnetic field at t₁, B₁ = 4.91 T/s

final magnitude of magnetic field at t₂, B₂ = 5.42 T/s

The magnitude of the induced emf in the coil is given by;

[tex]E = -N\frac{\delta \phi}{\delta t} \\\\E =-N (\frac{\delta B}{\delta t} )A\\\\E = -NA(\frac{B_1-B_2)}{\delta t} \\\\E = NA(\frac{B_2-B_1)}{\delta t} \\\\E = 20(0.0015)(5.42-4.91)\\\\E = 0.0153 \ V\\\\E = 15.3 \ mV[/tex]

Therefore, the magnitude of the induced emf in the coil is 15.3 mV

Answer:

The unit of work is joules

Force and displacement are used to calculate the work done by an object. This work is measured in the units of Joules. Thus, the correct option is A.

Work can be defined as the force that is applied on an object which shows some displacement. Examples of work done include lifting an object against the Earth's gravitational force, and driving a car up on a hill. Work is a form of energy. It is a vector quantity as it has both the direction as well as the magnitude. The standard unit of work done is the joule (J). This unit is equivalent to a newton-meter (N·m).

The nature of work done by an object can be categorized into three different classes. These classes are positive work, negative work and zero work. The nature of work done depends on the angle between the force and displacement of the object. Positive work is done if the applied force displaces the object in its direction, then the work done is known as positive work. Negative work is opposite of positive work as in this work, the applied force and displacement of the object are in opposite directions to each other and zero work is done when there is no displacement.

Therefore, the correct option is A.

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Explanation:

Let the first wave is :

[tex]y_1=A\sin\omega t[/tex]

And another wave is :

[tex]y_2=A\sin (\omega t+\phi)[/tex]

[tex]\phi[/tex] is phase difference between waves

Let y is the resultant of these two waves. So,

[tex]y =y_1+y_2[/tex]

The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,

[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]

So, the phase difference between the two waves is [tex]\pi[/tex].

Answer: The width is 1.25692 μm

Explanation:

The data that we have here is:

Distance between the single slit to the screen = L = 43cm

λ = 636 nm

Distance from the central maximum to the first minimum = Z =  3.8mm

We know that the angle for the destructive diffraction is:

θ = pλ/a

where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,

then we have:

θ = (636nm/a)

And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:

Tg(θ) =  Z/L

Tan(636nm/a) = 3.8cm/43cm

First, we need to use the same units in the right side:

3.8mm = 0.38cm

Tg(636nm/a)  = 0.38cm/43cm

636nm/a = Atg(  0.38cm/43cm ) = 0.506

a = 636nm/0.506 = 1,256.92 nm

1 μm = 1000nm

then:

a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm

Answer:

D. Metallic atoms have valence shells that are mostly empty, which

means these atoms are more likely to give up electrons and allow

them to move freely.

Explanation:

Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.

In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.

Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.

Complete question:

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.

Answer:

The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Explanation:

Given;

length of solenoid, L = 2.4 m

radius of solenoid, R = 1.7 cm = 0.017 m

current in the solenoid, I = 0.19 A

number of turns of the solenoid, N = 2100 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μnI

Where;

μ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length = N/L

I is current in the solenoid

B = μnI = μ(N/L)I

B = 4π x 10⁻⁷(2100 / 2.4)0.19

B = 4π x 10⁻⁷ (875) 0.19

B = 2.089 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Answer:

La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.

Explanation:

Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:

[tex]v = v_{o} + g\cdot t[/tex]

[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]

Donde:

[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.

[tex]t[/tex] - Tiempo, medido en segundos.

[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.

[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.

Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:

[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]

[tex]v = -196.14\,\frac{m}{s}[/tex]

[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]

[tex]y-y_{o} = -1961.4\,m[/tex]

La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Answer:

a) 94.26 g/s

b) 4.713 cm/s

c) 2356.5 cm^2

Explanation:

a) velocity of blood through the aorta = 30 cm/s

radius of aorta = 1 cm

density of blood = 1 g/cm^3

Area of the aorta = [tex]\pi r^{2}[/tex] = 3.142 x [tex]1^{2}[/tex] = 3.142 cm^2

Flow rate through the aorta Q = AV

where A is the area of aorta

V is the velocity of blood through the aorta

Q = 3.142 x 30 = 94.26 cm^3/s

Current of blood through aorta [tex]I[/tex] = Qρ

where ρ is the density of blood

[tex]I[/tex] = 94.26 x 1 = 94.26 g/s

b) Velocity of blood in the major aorta = 30 cm/s

Area of the aorta = 3.142 cm^2

Velocity of blood in the major arteries = ?

Area of major arteries = 20 cm^2

From continuity equation

[tex]A_{ao} V_{ao} = A_{ar} V_{ar}[/tex]

where

[tex]V_{ao}[/tex] = velocity of blood in the major arteries

[tex]A_{ao}[/tex] = Area of the aorta

[tex]V_{ar}[/tex] = velocity of blood in the major arteries

[tex]A_{ar}[/tex] = Area of major arteries

substituting values, we have

3.142 x 30 = 20[tex]V_{ar}[/tex]

94.26 = 20[tex]V_{ar}[/tex]

[tex]V_{ar}[/tex]  = 94.26/20 = 4.713 cm/s

c) From continuity equation

[tex]A_{ar} V_{ar} = A_{c} V_{c}[/tex]

where

[tex]A_{ar}[/tex] = Area of major arteries = 20 cm/s

[tex]V_{ar}[/tex] = velocity of blood in the major arteries = 4.713 cm/s

[tex]A_{c}[/tex] = Area of the capillary system = ?

[tex]V_{c}[/tex] = velocity of blood in the capillary system = 0.04 cm/s

substituting values, we have

20 x 4.713 = [tex]A_{c}[/tex]  x 0.04

94.26 = 0.04[tex]A_{c}[/tex]

[tex]A_{c}[/tex]  = 94.26/0.04 = 2356.5 cm^2

This question involves the concepts of volumetric flow rate, continuity equation, and flow velocity.

a) Total current of the blood passing through the aorta is "94.2 g/s".

b) The velocity of blood in major arteries is "4.71 cm/s".

c) The cross-sectional area of the capillary system is "2356.2 cm²".

a)

First, we will find the volumetric flow rate of the blood, using the continuity equation's formula:

[tex]Q=Av[/tex]

where,

Q = volumetric flow rate = ?

A = cross-sectional area of aorta

A =  [tex]\pi(r)^2=\pi(1\ cm)^2= 3.14\ cm^2[/tex]

v = flow velocity = 30 cm/s

Therefore,

[tex]Q=(3.14\ cm^2)(30\ cm/s)[/tex]

Q = 94.25 cm³/s

Now, the blood current will be given as:

I = Qρ

where,

I = current = ?

ρ = blood density = 1 g/cm³

Therefore,

I = (94.2 cm³/s)(1 g/cm³)

I = 94.2 g/s

b)

Now, this volumetric flow rate will be constant in major arteries:

[tex]Q = A_r v_r\\\\v_r=\frac{Q}{A_r}[/tex]

where,

Ar = cross-section area of major arteries = 20 cm²

vr = flow velocity of blood in major arteries = ?

Therefore,

[tex]v_r=\frac{94.25\ cm^3/s}{20\ cm^2}[/tex]

vr = 4.71 cm/s

c)

Now, this volumetric flow rate will be constant in capillaries:

[tex]Q = A_c v_c\\\\A_c=\frac{Q}{v_c}[/tex]

where,

Ac = cross-section area of capillaries = ?

vc = flow velocity of blood in capillaries = 0.04 cm/s

Therefore,

[tex]A_c=\frac{94.25\ cm^3/s}{0.04\ cm/s}[/tex]

Ac = 2356.2 cm²

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Answer:

The  initial velocity is  [tex]v_h = 8.66 \ m/s[/tex]

Explanation:

From the question we are told that

    The height of the tree is  [tex]h = 3.30\ m[/tex]

    The distance of the position of landing from base  is  [tex]d = 5.30 \ m[/tex]

According to the second equation of motion

    [tex]h = u_o * t + \frac{1}{2} at^2[/tex]

[tex]Where\ u_o[/tex] is the initial velocity in the vertical axis  

           a  is equivalent to acceleration due to gravity which is positive because the tiger is downward

    So

     [tex]3 = 0 + 0.5 * 9.8 *t^2[/tex]

=>    [tex]t = \frac{3 }{9.8 * 0.5}[/tex]

      [tex]t = 0.6122\ s[/tex]

Now the initial velocity in the horizontal direction is mathematically evaluated as

         [tex]v_h = \frac{5.30}{0.6122}[/tex]

        [tex]v_h = 8.66 \ m/s[/tex]

Answer:

C) proton acceptor

Explanation:

A proton acceptor is another name for a base,

Answer:

The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s

Explanation:

I'll assume that the correct question is

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?

mass of box = 51 kg

for the first 12 m, it is pulled with a constant force of 240 N

The acceleration of the box for this first 12 m will be

from F = ma

a = F/m

where F is the pulling force

m is the mass of the box

a is the acceleration of the box

a = 240/51 = 4.71 m/s^2

Since the body started from rest, the initial velocity u = 0

applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have

[tex]v^{2}= u^{2}+2as[/tex]

where v is the final velocity

u is the initial velocity which is zero

a is the acceleration of 4.71 m/s^2

s is the distance covered which is 12 m

substituting value, we have

[tex]v^{2}[/tex] = 0 + 2(4.71 x 12)

[tex]v^{2}[/tex]  = 113.04

[tex]v = \sqrt{113.04}[/tex] = 10.63 m/s

For the final 10.5 m, coefficient of friction is 0.21

from  f = μF

where f is the frictional force,

μ is the coefficient of friction = 0.21

and F is the pulling force of the box 240 N

f = 0.21 x 240 = 50.4 N

Net force on the box = 240 - 50.4 = 189.6 N

acceleration = F/m = 189.6/51 = 3.72 m/s^2

Applying newton's equation of motion

[tex]v^{2}= u^{2}+2as[/tex]

u is initial velocity, which in this case =  10.63 m/s

a = 3.72 m/s^2

s = 10.5 m

v = ?

substituting values, we have

[tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)

[tex]v^{2}[/tex]  = 112.9 + 78.12

v  = [tex]\sqrt{191.02}[/tex]  = 13.82 m/s