A man pushes a 25kg box up an incline 2.0m by applying a steady force of 95N parallel to the incline. The box moves up the incline at a steady speed. The incline makes an angle 15 degrees to the horizontal

a) What is the force of friction between the incline and the box

b)If the box is released at the top of the incline, what will its speed be at the bottom

Answer:

0.1575 m/s^2

Explanation:

Solution:-

- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).

- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.

- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).

- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:

                         [tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]

- We will plug in the given parameters in the equation of motion given above:

                         [tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]

Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2

Answer:

The mass of the cargo is [tex]M = 188.43 \ kg[/tex]

Explanation:

From the question we are told that

    The radius of the spherical balloon is  [tex]r = 7.40 \ m[/tex]

     The mass of the balloon is  [tex]m = 990\ kg[/tex]  

The volume of the spherical balloon is mathematically represented as

     [tex]V = \frac{4}{3} * \pi r^3[/tex]

substituting values

      [tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]

      [tex]V = 1697.6 \ m^3[/tex]

The total mass  the balloon can lift is mathematically represented as

     [tex]m = V (\rho_h - \rho_a)[/tex]

where [tex]\rho_h[/tex] is the density of helium with a  value of

       [tex]\rho_h = 0.179 \ kg /m^3[/tex]

and  [tex]\rho_a[/tex] is the density of air with a value of

        [tex]\rho_ a = 1.29 \ kg / m^3[/tex]

substituting values

          [tex]m = 1697.6 ( 1.29 - 0.179)[/tex]

         [tex]m = 1886.0 \ kg[/tex]

Now the mass of the cargo is mathematically evaluated as

        [tex]M = 1886.0 - 1697.6[/tex]

        [tex]M = 188.43 \ kg[/tex]

Answer:

The velocity is  [tex]v_2= 0.45 \ m/s[/tex]

Explanation:

From the question we are told that

      The initial speed of the hot water is  [tex]v_1 = 0.85 \ m/s[/tex]

     The pressure from the heater  [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]

      The height of the hot water before flowing is  [tex]h_1 = 0 \ m[/tex]

      The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]

       The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]

       The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]

Apply Bernoulli equation

      [tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]

Substituting values

     [tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]

=>   [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]

=>   [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]

=>    [tex]v_2= 0.45 \ m/s[/tex]

Answer:

Cube temperature = 526.83 K

Explanation:

Volume of the cube and sphere will be the same.

Now, volume of cube = a³

And ,volume of sphere = (4/3)πr³

Thus;

a³ = (4/3)πr³

a³ = 4.1187r³

Taking cube root of both sides gives;

a = 1.6119r

Formula for surface area of sphere is;

As = 4πr²

Also,formula for surface area of cube is; Ac = 6a²

Thus, since a = 1.6119r,

Then, Ac = 6(1.6119r)²

Ac = 15.5893r²

The formula for radiant power is;

Q' = eσT⁴A

Where;

e is emissivity

σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k

T is temperate in kelvin

A is Area

So, for the cube;

(Qc)' = eσ(Tc)⁴(Ac)

For the sphere;

(Qs)' = eσ(Ts)⁴(As)

We are told (Qc)' = (Qs)'

Thus;

eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)

eσ will cancel out to give;

(Tc)⁴(Ac) = (Ts)⁴(As)

Since we want to find the cube's temperature Tc,

(Tc)⁴ = [(Ts)⁴(As)]/Ac

Plugging in relevant figures, we have;

(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²

r² will cancel out to give;

(Tc)⁴ = [556⁴ × 4π]/15.5893

Tc = ∜([556⁴ × 4π]/15.5893)

Tc = 526.83 K

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

Answer:

a. 42N

b. 11.8m/s

c. 1.69s

d. 160N

Explanation:

a)  The tension of the rope is 130.66 N.

b) The speed of the bucket while strike the water = 4.64 m/s.

c) The time of fall is  = 4.303 second.

d)  While the bucket is falling, what is the force exerted on the cylinder by the axle is  130.66 N.

Mass of the water bucket; M = 15.0 kg

Mass of the cylinder; m =  12.0 kg

Height of the bucket; h = 10.0 m.

They are connected by a rope and a pivots.

So, acceleration of them is same and let it be a.

So equation of motion of both of them be:

Mg - T = Ma

and, T - mg = ma

Hence, a = g(M-m)/(M+m)

= 9.8(15-12)/(15+12)

= 1.08 m/s²

And, T = m(g+a)

= 12.0(9.8+1.08)

= 130.66 N.

a) so tension of the rope is 130.66 N.

b) speed of the bucket while strike the water = √2ah =√(2×1.08×10.0) m/s = 4.64 m/s.

c) The time of fall is = √2h/a = √(2×10/1.08) second = 4.303 second.

d) While the bucket is falling, what is the force exerted on the cylinder by the axle is tension of the rope, that is, 130.66 N.

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Answer:

Explanation:

a )

Radius of the sun = .69645 x 10⁹ m .

600 times = 600 x .69645 x 10⁹ m

= 4.1787 x 10¹¹ m .

surface area A = 4π (4.1787 x 10¹¹)²

= 219.317 x 10²²

energy radiated E = σ A Τ⁴

= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴

= 100695 x 10²⁶ J

To know the wavelength of photon emitted

[tex]\lambda_mT= b[/tex]

[tex]\lambda_m= \frac{b}{T}[/tex]

= 2.89777 x 10⁻³ / 3000

= 966 nm

= 1275 /966 eV

1.32 x 1.6 x 10⁻¹⁹ J

= 2.112 x 10⁻¹⁹ J

No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹

= 47677.5 x 10⁴⁵

= .476 x 10⁵⁰ .

b )

energy radiated by our sun per second

E₂ = σ A 5800⁴

energy radiated by Betelgeuse per second

E₁ = σ  x 600²A x  3000⁴

E₁ / E₂  = σ  x 600²A x  3000⁴ / σ A 5800⁴

= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸

= 25.76 x 10⁸ x 10⁻⁵

= 25760 times .

Answer:

Explanation:

The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]

The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.

Resistors are in series whenever the current flows through the resistors sequentially. It is given by

[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]

Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.

The equivalent resistance is the total resistance measured in a parallel. It is given by

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]

Given:

Resistor, [tex]R_{1} = 100 ohm[/tex]

Resistor, [tex]R_{2} = 200 ohm[/tex]

Voltage, [tex]V = 10 Volt[/tex]

Since, resistors are connected in parallel, the equivalent resistor is given by,

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]

[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]

[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]

Hence, the equivalent resistor is [tex]66.66 ohm[/tex].

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Answer:

The value of m is 2635.294 grams.

Explanation:

Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:

[tex]f = \frac{\omega}{2\pi}[/tex]

Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.

For a mass-spring system under simple harmonic motion, the angular frequency is:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

Where:

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m[/tex] - Mass, measured in kilograms.

The following equation is obtained after replacing angular frequency in frequency formula:

[tex]f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }[/tex]

As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:

[tex]f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}[/tex]

If [tex]m_{2} = m_{1} + 700\,g[/tex], [tex]f_{1} = 0.72\,hz[/tex] and [tex]f_{2} = 0.64\,hz[/tex], the resulting expression is simplified and then initial mass is found after clearing it:

[tex]f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}[/tex]

[tex]f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)[/tex]

[tex]\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g[/tex]

[tex]\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g[/tex]

[tex]m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}[/tex]

[tex]m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}[/tex]

[tex]m_{1} = 2635.294\,g[/tex]

The value of m is 2635.294 grams.

Answer:

The correct option is;

- 4x

Explanation:

From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source

The inverse square law can be presented as follows;

[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]

As the distance, r, increases, the surface it covers also increases by the power of 2

Therefore, where the distance increases from r to 2·r, we have;

When, I, remain constant

[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]

The surface increases to 4·S by the inverse square law

Therefore, the correct option is 4 × x.

An object's resistance to any change in motion is the Inertia of the object.

Answer:

Final Temperature = 71 °C

Explanation:

In this case, the ideal gas equation is written as;

PV = mRT

Where;

P is pressure

V is volume

m is mass

R is gas constant

T is temperature

We will take the volume to be constant.

So, in the initial state, we have;

P1•V = m1•R•T1 - - - eq(1)

In the final state, we have;

P2•V = m2•R•T2 - - - - eq(2)

Combining eq (1) and eq(2),we have;

P1•m2•R•T2 = P2•m1•R•T1

Dividing both sides by R gives;

P1•m2•T2 = P2•m1•T1

Making T2 the subject gives;

T2 = (P2•m1•T1)/(P1•m2)

Now, we are given;

m1 = 2kg

m2 = ½*2 = 1kg

P1 = 4 atm

P2 = 2.2 atm

T1 = 40°C = 273 + 40 K = 313K

Plugging in this values into the T2 equation, we have;

T2 = (2.2 × 2 × 313)/(4 × 1)

T2 = 344 K

Converting to °C, we have;

T2 = 344 - 273 = 71 °C

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,

Distance = 40 meters

Time = 5 seconds

Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec

Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,

Distance = 64 meters

Time = 8 seconds

Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec

Hence, During their periods of travel, the cars definitely had the same velocity.

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Answer:

it takes the shape and size of the container that it is in

Explanation:

Answer:

it takes the shape and size of a container

Answer:

C = 44.75 x 10⁻⁸ F

Explanation:

Assuming no loss of energy between capacitor and inductor

energy in inductor initially = 1/2 Li₀² where L is inductance and i₀ is peak current .

= .5 x 6 x 10⁻³ x .57²

= .97 x 10⁻³ J .

This energy is transferred to capacitor .

energy of capacitor = 1/2 CV²

= .5 x C x 66²

= 2178 C

2178C = .97 x 10⁻³

C = 44.75 x 10⁻⁸ F .

The magnetic energy stored in the inductor is transformed into electrical energy stored in the capacitor. The value of capacitance for the given circuit is 44.75 x 10⁻⁸ F

According to the law of conservation of energy, the magnetic energy stored in the inductor will be gradually lost and this energy will be stored in the capacitor as electrical energy.

Initially, the energy in the inductor is:

E = 1/2 Li₀²

where L is inductance

and i₀ is peak current.

E = 0.5 × 6 × 10⁻³ × (0.57)²

E = 0.97 × 10⁻³J

This energy is transformed into electrical energy stored in the capacitor.

So the capacitor energy is:

E = 1/2 CV²

where C is the capacitance

E = 0.5 × C × 66²

E = 2178 C

0.97 x 10⁻³ = 2178 C

C = 44.75 x 10⁻⁸ F

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Answer:

E = 3.45*10^-19 N/C

Explanation:

a) The electric field between two parallel plates id given by the following formula:

[tex]E=\frac{\sigma}{\epsilon_o}[/tex]           (1)

where:

σ: surface charge density of the plates = 39.0nC/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2

You replace these values in the equation (1):

[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]

The electric field in between the parallel plates is 3.45*10^-19 N/C

Answer:

[tex]v_{i}=10.10 m/s[/tex]

Explanation:

The equation of the position is:

[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]

Where:

v(i) is the initial velocity

The initial position y(i) will be zero and the final position y = 3.60 m.

So, we just need to solve this equation for v(i).

[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]

[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]

[tex]v_{i}=10.10 m/s[/tex]

Therefore, the initial velocity is 10.10 m/s upwards.

I hope it helps you!

Answer: B

Explanation:

I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges

Answer:B :)

Explanation:

Answer:

0.981 rad/sec^2

Explanation:

mass that pulls on string = 5 kg

weight due to mass = 5 x 9.81 = 49.05 N

radius of rod = 0.1 m

torque produced by this force on the rod = force x radius

torque = 49.05 x 0.1 = 4.905 N-m

mass of disk = 125 kg

radius of disk = 0.2 m

moment of inertia of the disk I = m[tex]r^{2}[/tex]

I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2

from the equation, T = Iα

where T is torque

I is moment of inertia

α is angular acceleration

imputing values,

4.905 = 5α

α = 4.905/5 = 0.981 rad/sec^2

Answer:

The  drag coefficient is  [tex]D_z = 1.30512[/tex]  

Explanation:

From the question we are told that

     The density of air is  [tex]\rho_a = 1.21 \ kg/m^3[/tex]

     The diameter of bottom part is  [tex]d = 0.15 \ m[/tex]

The  power trend-line  equation is mathematically represented as

      [tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]

let assume that the velocity is  20 m/s

Then

      [tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]

       [tex]F_{\alpha } = 5.1453 \ N[/tex]

The drag coefficient is mathematically represented as

      [tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]

Where  

     [tex]F_{\alpha }[/tex] is the drag force

      [tex]\rho[/tex] is the density of the fluid

       [tex]v[/tex] is the flow velocity

       A is the area which mathematically evaluated as

       [tex]A = \pi r^2 = \pi \frac{d^2}{4}[/tex]

substituting values

     [tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]

     [tex]A = 0.0176 \ m^2[/tex]

Then

   [tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]

   [tex]D_z = 1.30512[/tex]  

Answer:

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Explanation:

The Free Body Diagram associated with the experiment is presented as attachment included below.

Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.

Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:

[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]

[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]

Where:

[tex]W[/tex] - Weight of the eraser, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force, measured in newtons.

[tex]\theta[/tex] - Angle of the incline, measured in degrees.

The maximum allowable static friction force is:

[tex]f = \mu_{s} \cdot N[/tex]

Where:

[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

Likewise, the kinetic friction force is described by the following model:

[tex]f = \mu_{k} \cdot N[/tex]

Where:

[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])

In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:

[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]

[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]

But [tex]N = m\cdot g \cos \theta[/tex], so that:

[tex]\mu = \tan \theta[/tex]

Now, coefficients of static and kinetic friction are, respectively:

[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]

[tex]\mu_{s} \approx 0.716[/tex]

[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]

[tex]\mu_{k} \approx 0.596[/tex]

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Answer:

It's true

Explanation:

Because the ship is mafe up of aluminium, which is a light metal.

Answer:

False

Explanation:

Took The Quiz

Answer:

Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL

AAAAAAAAAAAA is the answer

Answer:

18 N

Explanation:

Force can be found using the following formula.

f= m*a

where m is the mass and a is the acceleration.

We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.

m= 36 kg

a= 0.5 m/s^2

Substitute these values into the formula.

f= 36 kg * 0.5 m/s^2

Multiply 36 and 0.5

f=18 kg m/s^2

1 kg m/s^2 is equivalent to 1 Newton, or N.

f= 18 Newtons

The force being applied is 18 kg m/s^2, Newtons, or N

Answer:If a wave y(x, t) = (6.0 mm) sin(kx + (600 rad/s)t + Φ) travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?

Explanation:

Answer:

B. revolve around the sun

   rotate on an axis

   generally have moons

Explanation:

edge 2021

Answer:

hypothesis , hope it helps

Explanation:

Answer:

Inference

Explanation:

Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.

Answer:

the correct answer is B

Explanation:

We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.

For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.

Let's write the kinematic equation for the two bodies

The block. At the highest point of the path

      y = - ½ g t2

The ball, in its vertical movement

     y = vo t - ½ g t2

therefore the correct answer is B