In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 88 kgkg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 NN .
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Answer:

About 6.26m/s

Explanation:

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Divide both sides by mass:

[tex]gh=\dfrac{1}{2}v^2[/tex]

Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.

[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]

Hope this helps!

The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Given data:

The height of vertical cliff is, h = 4.0 m.

Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,

Kinetic energy = Gravitational potential energy

[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]

Here,

m is the mass of rock.

v is the speed of rock.

g is the gravitational acceleration.

Solving as,

[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]

Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

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Answer: C. Petal. Length

Explanation: Petal are unit of Corolla which are usually brightly colored. This part of a plant or flower, helps attracts insects to the plant for pollination. And also provide protection to the reproductive parts of the plant or flower.

Examples of flowers with petals is the Sun Flower, which coincidentally is the flower plant with most petals.

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              [tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             [tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             [tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          [tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex]  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         [tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]

                          [tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]

Answer:1m/s

Explanation: As the stationary ball is hit by the moving ball ,the two moves together after collision, with a single velocity. The attached photo further explains how the answer is calculated

When stationary ball is hit by the moving ball, both the balls moves together after collision. The final velocity of the object after collision is 1 m/s.

When stationary ball is hit by the moving ball, both the balls moves together after collision

The conservation of momentum,

[tex]\bold {m_1 u_1 + m_2u_2 = (m_1+m_2) V}\\[/tex]

Where,

m1 - initial mass = 20 kg

m2 - final mass =10 kg

u1 - initial velocity = 0 m/s (object at rest)

u2 - final velocity = 3 m/s

V- velocity after collision = ?

Put the values int he formula and calculate for V2,

[tex]\bold { 10 \times 0 + 20 \times 3 = (10+20) V}\\\\\bold {V = \dfrac {30}{30}}\\\\\bold {V = 1\ m/s}[/tex]

Therefore, final velocity of the object after collision is 1 m/s.

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Answer:

v =   11 m/s   is her final speed

Explanation:

work done by gravity = m g Δh =   40×9.8×10   = 3920 Joules

Work done by friction = - force×distance =   - 20×100   =   - 2000 Joules

[minus sign because friction force is opposite to the direction of motion]

Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2   = 500 Joules

Now, by work energy theorem

Work done = change in kinetic energy.

Final K.E. = initial K.E. + total work =    500 + 3920 - 2000  = 2420 Joules

Now, Final K.E. = (1/2) m v^2  [final speed being v= speed at the bottom]

⇒  2420 = (1/2)×40×v^2

   ⇒  121 = v^ 2

  v =   11 m/s   is her final speed

Answer:

a) WT = 137.5 J

b) v2 = 2.34 m/s

Explanation:

a) The total work done on the block is given by the following formula:

[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex]          (1)

Fp: force parallel to the displacement of the block = 150N

Ff: friction force

d: distance = 5.0 m

Then, you first calculate the friction force by using the following relation:

[tex]F_f=\mu_k N=\mu_k Mg[/tex]        (2)

μk: coefficient of kinetic friction = 0.25

M: mass of the block = 50kg

g: gravitational constant = 9.8 m/s^2

Next, you replace the equation (2) into the equation (1) and solve for WT:

[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]

The work done over the block is 137.5 J

b) If the block started from rest, you can use the following equation to calculate the final speed of the block:

[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex]     (3)

WT: total work = 137.5 J

v2: final speed = ?

v1: initial speed of the block = 0m/s

You solve the equation (3) for v2:

[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]

The final speed of the block is 2.34 m/s

Answer:

visual basics

Explanation:

not suited for programming, slower than the other languages. hard to translate to other operating systems

Answer and Explanation: The types of reaction represented in the question are the following: S is Synthesis; D is Decomposition; SD is Single Displacement; and DD is Double Displacement.

The reaction showed is: HI ⇄ H₂ + I₂

This reaction's type is Decomposition, because it is a reaction in which a compund breaks down into 2 or more simpler compounds. Most of them requires energy that could be in form of heat, light or electricity.

To balance a chemical equation, remember that, according to the Conservation of Mass Law, matter can not be created or destroyed just transformed. Then, the amount of reagents that enters the reaction has to be same amount that reacts to become the products, i.e.:

For the decomposition reaction, there are 2 atoms of hydrogen and 2 atoms of iodine being produced while only 1 atom of each element enters the reaction.

The balanced reaction will be: 2HI ⇄ H₂ + I₂

Answer:

12 m/s²

Explanation:

Maximum acceleration is a = Aω², where A is the amplitude and ω is the frequency.

a = (0.0030 m) (10 rev/s × 2π rad/rev)²

a = 12 m/s²

Answer:

The angular velocity is [tex]w_f = 4.503 \ rad/s[/tex]

Explanation:

From the question we are told that

   The mass of the child is  [tex]m_c = 46.2 \ kg[/tex]

    The radius of the merry go round is  [tex]r = 1.9 \ m[/tex]

     The moment of inertia of the merry go round is [tex]I_m = 130.09 \ kg \cdot m^2[/tex]

      The angular velocity of the merry-go round is  [tex]w = 2.4 \ rad/s[/tex]

       The position of the child from the center of the merry-go-round is  [tex]x = 0.779 \ m[/tex]

According to the law of angular momentum conservation

    The initial angular momentum  =  final  angular momentum

So  

       [tex]L_i = L_f[/tex]

=>     [tex]I_i w_i = I_fw_f[/tex]

Now   [tex]I_i[/tex] is the initial moment of inertia of the system which is mathematically represented as

          [tex]I_i = I_m + I_{b_1}[/tex]

Where  [tex]I_{b_i}[/tex] is the initial moment of inertia of the boy which is mathematically evaluated as

      [tex]I_{b_i} = m_c * r[/tex]

substituting values

      [tex]I_{b_i} = 46.2 * 1.9^2[/tex]

      [tex]I_{b_i} = 166.8 \ kg \cdot m^2[/tex]

Thus

   [tex]I_i =130.09 + 166.8[/tex]        

   [tex]I_i = 296.9 \ kg \cdot m^2[/tex]      

Thus  

     [tex]I_i * w_i =L_i= 296.9 * 2.4[/tex]

       [tex]L_i = 712.5 \ kg \cdot m^2/s[/tex]

Now  

     [tex]I_f = I_m + I_{b_f }[/tex]

Where  [tex]I_{b_f}[/tex] is the final  moment of inertia of the boy which is mathematically evaluated as

         [tex]I_{b_f} = m_c * x[/tex]

substituting values

         [tex]I_{b_f} = 46.2 * 0.779^2[/tex]

         [tex]I_{b_f} = 28.03 kg \cdot m^2[/tex]

Thus

      [tex]I_f = 130.09 + 28.03[/tex]

      [tex]I_f = 158.12 \ kg \ m^2[/tex]

Thus

     [tex]L_f = 158.12 * w_f[/tex]

Hence

      [tex]712.5 = 158.12 * w_f[/tex]

       [tex]w_f = 4.503 \ rad/s[/tex]

Answer:

(b) Torque will increase.

Explanation:

Torque is given as the product of force and moment arm (radius).

τ = F x r

F = τ / r

where;

F is force

τ  is torque

r is radius (moment arm)

Keeping force constant, we will have the following;

τ ∝ r

This shows that torque is directly proportional moment arm (radius), thus increase in moment arm, will cause increase in torque.

For instance;

let the constant force = 5 N

let the initial moment arm, r = 2m

Torque, τ  = 5 N x 2m = 10 Nm

When the moment arm is increased to 4 m

Torque, τ  = 5 N x 4m = 20 Nm

Therefore, at a constant force, increasing in the Moment arm, will cause increase in torque.

Coorect option is "(b) Torque will increase."

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm  - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

Final Length = 30 cm

Answer:

F = 22.75 lb

μ₁ = 0.15

Explanation:

The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,

F = μR

F = μW

where,

F = Smallest force needed to move dresser = ?

μ = coefficient of static friction = 0.25

W = Weight of dresser = 91 lb

Therefore,

F = (0.25)(91 lb)

F = 22.75 lb

Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:

F = μ₁W₁

where,

μ₁ = coefficient of static friction between shoes and floor

W₁ = Weight of man = 151 lb

Therefore,

22.75 lb = μ₁ (151 lb)

μ₁ = 22.75 lb/151 lb

μ₁ = 0.15

Answer:

If the initial speed is doubled the time is also doubled

Explanation:

You have that a car with velocity v is decelerated by a constant acceleration a in a time t.

You use the following equation to establish the previous situation:

[tex]v'=v-at[/tex]     (1)

v': final speed of the car  = 0m/s

v: initial speed of the car

From the equation (1) you solve for t and obtain:

[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex]     (2)

To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:

[tex]v'=v_1-at'[/tex]     (3)

v' = 0m/s

v1: new initial speed = 2v

t': new time

You solve the equation (3) for t':

[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]

If the initial speed is doubled the time is also doubled

Answer:

Both the parts will weigh the same

Explanation:

Answer:

The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].

Explanation:

Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:

[tex]I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}[/tex]

Where:

[tex]I[/tex] - Moment of inertia of a disk, measured in kilogram-square meter.

[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed, measured in radians per second.

This relationship is simplified and final angular speed can be determined in terms of initial angular speed:

[tex]\omega_{f} = \frac{1}{2}\cdot \omega_{o}[/tex]

Given that [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the angular speed of the new system is:

[tex]\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)[/tex]

[tex]\omega_{f} = 3\,\frac{rad}{s}[/tex]

The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].

Answer:

Option B is the correct option.

Explanation:

The phenomenon of the curtain to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's principle,when the speed of moving fluid increases the pressure within the fluid decrease.When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more.So the curtain moves outside because of low pressure.

Hope this helps...

Good luck on your assignment...

Answer:

a) 5.19 W/m²

b) 1.79 J

Explanation:

For the calculation of intensity, I. We have

I = E(rms)² / (cμ), where

c = speed of light

μ = permeability of free space

I = (62.5 / √2)² / [(2.99 x 10^8) (1.26 x 10^-6)]

I = 1954 / 376.74

I = 5.19 W/m²

Therefore, the intensity, I = 5.19 W/m²

t = 14.9 s

A = 0.0231 m²

Amount if energy flowing, U = IAt

U = (5.19) (0.0231) (14.9) J

U = 1.79 J

Answer:

22.66Nm²/C

Explanation:

Flux through an electric field is expressed as ϕ = EAcosθ

When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.

From the formula above where:

EA = 25N.m^2/C

θ = 25°

ϕ = 25cos 25°

ϕ = 22.66Nm²/C

Answer:

756 N.

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this particular Question or problem effectively;

=> The bathroom scale in an elevator is not moving.

=> " The scale reads 588 N. "

=> "The elevator then moves down at an acceleration of - 2.8 m/s2."

For the second point above: 588 N = mass × gravity.

Thus, the scale that will read while the elevator moves down is given below;

(Mass × gravity) - (mass × acceleration).

(The value of scale when it is not moving) × (1 + acceleration/gravity).

=> 588 × { 1 + 2.8/9.8} = 756 N

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

Answer:

Explanation:

Given that,

The length of the beam L = 3.10m

The mass of the steam beam [tex]m_1[/tex] = 430kg

The mass of worker [tex]m_2[/tex] = 69.0kg

The distance from  the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]

and our length of beam is 3.10m

so the distance from  the fixed point to centre of gravity of beam is

[tex]\frac{3.10}{2}=1.55m[/tex]

Then the net torque is

[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]

[tex]W_s[/tex] is the weight of steel rod

[tex]=430\times9.8=4214N[/tex]

[tex]W_w[/tex] is the weight of the worker

[tex]=69\times9.8\\\\=676.2N[/tex]

Torque can now be calculated

[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]

≅ 9169Nm

Answer:

Negative charge with a magnitude of 5.9*10^-6C

Explanation:

In order to calculate the required charge, for the penny is able to rise into the air with the given acceleration, you take into account the following formula for the electric force and also the Newton second law:

[tex]F_E=qE=ma[/tex]            (1)

FE: electric force:

q: charge = ?

E: electric field = 100N/C

m: mass of the penny = 3.11g = 3.11*10^-3kg

a: acceleration of the charge = 0.190m/s^2

The positive charges experiences a force in the same direction of the electric field, and the negative charge, a force in a opposite direction.

The electric field points downward, then, the charge is negative (it is acceleration upwards).

Next, you solve the equation (1) for a, and replace the values of the other parameters:

[tex]q=\frac{ma}{E}=\frac{(3.11*10^{-3}kg)(0.190m/s^2)}{100N/C}=5.9*10^{-6}C[/tex]

The magnitude of the charge is 5.9*10^-6C and its sign is negative

Answer:

The correct answer is - 405 km/hr

Explanation:

given:

2-m- = diameter satellite dish

81-km/hr = Y

0.4-m- = diameter model dish.

Reynolds number is essential for this, thus,

Vm Dm/vm = VD/v

(D is the diameter of the dish)

Vm = (vm/v) (D/Dm) Y

and if vm/v = 1, then

Vm = (2/0.4) (81)

= 405 km/hr

Thus, the correct answer is - 405 km/hr

Answer:

D

Explanation:

Answer:

It is D

Explanation: No cap

Answer:

a)15m

b)6.25s

Explanation:

A) She is ½ a wavelength away, or

d = λ/2 = 30/2 = 15 m

B)Speed of the wave:

V=fλ = λ/T

so,

T=λ/V= 30/4.8

T=6.25s

a) The distance from the wall is 15m

b) The period of her up and down motion is 6.25s

a.

Since Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s .

Also,

She is ½ a wavelength away, or

d = λ/2

= 30/2

= 15 m

b)

Here the speed of wave should be used

T=λ/V

= 30/4.8

T=6.25s

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Answer:

1.603 s

Explanation:

Given that

Initial mass, = 0.45 kg

Initial period, = 1.45 s

Initial radius, = 0.14 m

Final mass, = 0.55 kg

Final period, = ?

Final radios, = 0.14 m

Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that

m₁r₁ω₁² = m₂r₂ω2²

Where, ω = 2π/T, on substituting, we have

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

The  rotation period for a mass of 0.550 kg at the same radius is 1.603 s

Since

Initial mass = 0.45 kg

Initial period = 1.45 s

Initial radius = 0.14 m

Final mass = 0.55 kg

Final radios = 0.14 m

Now the following formulas should be used.

m₁r₁ω₁² = m₂r₂ω2²

here ω = 2π/T

So,

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

hence, The  rotation period for a mass of 0.550 kg at the same radius is 1.603 s

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Answer:

a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]

Explanation:

a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

Where:

[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.

[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.

Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:

[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]

[tex]COP_{HP} = 14[/tex]

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]

The input work to deliver a determined amount of heat to the house:

[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]

If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:

[tex]W_{in} = \frac{3000\,J}{14}[/tex]

[tex]W_{in} = 214.286\,J[/tex]

b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

Where:

[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.

[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.

Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:

[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]

[tex]COP_{HP} = 7[/tex]

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]

The input work to deliver a determined amount of heat to the house:

[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]

If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:

[tex]W_{in} = \frac{3000\,J}{7}[/tex]

[tex]W_{in} = 428.571\,J[/tex]

Answer:

the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.

the book it is observed in all directions so that its reflection has to be diffused

Explanation:

The ray of light emitted by a candle is the light generated by the temperature of the flame, which is made up of the emissions of a black body at this temperature plus the emissions of the chemical elements that make up the candle.

The Light reflected from the cover of a book is the same incident light spectrum minus the wavelengths that create transitions in the elements of the cover, these wavelengths will be seen as dark areas.

As a consequence of the above, the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.

For the cover of the book form a specular reflection the incident rays are reflected in one direction and the rest would be dark, but in the book it is observed in all directions so that its reflection has to be diffused

Answer:

There is no difference between a light ray emitted by a candle flame and one reflected off the cover of a book. The candle flame is a source of light rays, but those rays could travel to the book cover and reflect off the book cover diffusely so that the same ray is actually emitted by the candle flame and reflected by the book. The reflection off the book is diffuse reflection because the book is visible from any angle.

Explanation: