the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is

Answers

Answer 1

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².


Related Questions

Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies

Answers

Explanation:

https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d

In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.

Answers

Answer:

c)  N_s / N_p = 115.15

Explanation:

Let's look for the voltage in the secondary, they do not indicate the power dissipated

          P = V_s i

          V_s = P / i

          V_s = 76 / 5.5 10⁻³

          V_s = 13.818 10³ V

the relationship between the primary and secondary of a transformer is

           [tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]

           [tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]

           Ns / Np = 13,818 10³ /120

           N_s / N_p = 115.15

The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field strength in volts per meter

Answers

Answer:

7.60× 10^6 V/m

Explanation:

electric field strength can be determined as ratio of potential drop and distance, I.e

E=V/d

Where E= electric field

V= potential drop= 74.0 mV= 0.07 V

d= distance= 9.20 nm = 9.2×10^-9 m

Substitute the values

E= 0.07/ 9.2×10^-9

= 7.60× 10^6 V/m

Based on the information in the table, what
is the acceleration of this object?

t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2

Answers

Answer:

Option A. –5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (v₁) = 9 m/s

Initial time (t₁) = 0 s

Final velocity (v₂) = –6 m/s

Final time (t₂) = 3 s

Acceleration (a) =?

Next, we shall determine the change in the velocity and time. This can be obtained as follow:

For velocity:

Initial velocity (v₁) = 9 m/s

Final velocity (v₂) = –6 m/s

Change in velocity (Δv) =?

ΔV = v₂ – v₁

ΔV = –6 – 9

ΔV = –15 m/s

For time:

Initial time (t₁) = 0 s

Final time (t₂) = 3 s

Change in time (Δt) =?

Δt = t₂ – t₁

Δt = 3 – 0

Δt = 3 s

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Change in velocity (Δv) = –15 m/s

Change in time (Δt) = 3 s

Acceleration (a) =?

a = Δv / Δt

a = –15 / 3

a = –5 m/s²

Thus, the acceleration of the object is

–5 m/s².

a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?

Answers

Charge me and do I name for meters

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits

Answers

Answer:

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Explanation:

Given :

The first dark fringe is for m = 0

[tex]$\theta_1 = \pm 19^\circ$[/tex]

Now we know for a double slit experiments , the position of the dark fringes is give by :

[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]

The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :

[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex]     (since, m = 0)

[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex]  or 1 : 1.54

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?

Answers

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?

Answers

Answer:

Power = Energy/time

Energy = Power xtime.

Time= 20hrs

Power = 100Watt =0.1Kw

Energy = 0.1 x 20 = 2Kwhr.

This Answer is in Kilowatt-hour ...

If the one given to you is in Joules

You'd have to Change your time to seconds

Then Multiply it by the power of 100Watts.

Define relative density.​

Answers

Relative density is the ratio of the density of a substance to the density of a given material.

A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would result in an induced emf in the loop?
A. Moving the loop outside of the magnetic field region.
B. Change the diameter of the loop.
C. Change the magnitude of the magnetic field.
D. Spin the loop such that its axis does not consistently line up with the magnetic field direction.

Answers

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

[tex]emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta[/tex]

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

Considering option A, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

Considering option B, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, considering option D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will change the angle between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

What are stepdown transformers used for

Answers

Answer:

Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.

Explanation:

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Thanks a lot

A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound

Answers

Answer:

64.20m

Explanation:

As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.

[tex]a^{2} +b^{2} = c^{2}[/tex]

where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x

[tex]51^{2} +39^{2} =x^{2}[/tex]

2,601 + 1,521 = [tex]x^{2}[/tex]

4,122 = [tex]x^{2}[/tex]   ... square root both sides

64.20 = x

Finally, we see that the shortest distance is 64.20m

A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate

(i) The speed of the train

(ii) The time it will taketo cover 240 m.​

Answers

Answer:

(I)

[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]

(ii)

[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]

A boy is playing with a water hose, which has an exit area of
10 cm2 and has water flowing at a rate of 2 m/s. If he covers
the opening of the hose with his thumb so that it now has an
open area of 2 cm2, what will be the new exit velocity of the
water?

Answers

Answer:

The exit velocity of water is  B. 15 m/s.

Explanation:

According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.

If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity  is v₂, then,

Rewrite the expression for v₂.

Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.

The exit speed of water from the hose is 15 m/s.

. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration  

              ∑ F = m a

in this case there are two forces on the x axis

             F_applied - fr = 0

since they indicate that the velocity is constant, consequently

             F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops

Answers

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations

Answers

Answer:

The height is the same

Explanation:

Because they were at the same height but they fell at different velocities

A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster?

Answers

Answer:

When an object is dropped, the "principal" force that acts on that object is the gravitational force.

Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:

g = 9.8m/s^2

So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)

Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.

But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)

And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)

Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.

If an object of a constant mass experiences a constant net force, it will have a constant what?

Answers

Explanation:

hope it helps !!!!!!!!!!!!!

If an object of a constant mass experiences a constant net force, it will have a constant acceleration.

What is force?

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

According to Newton's second law of motion:

Applied force = mass × acceleration.

Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.

Learn more about force here:

https://brainly.com/question/26115859

#SPJ6

4. Paper is solid in packets labelled 80 g/m2. This means that a sheet of paper of area
10 000cm? has a mass of 80 g. The thickness of each sheet is 0.11mm. What is the
density of the paper?
A 0.073 g/cm?
B 0.088 g/cm
C 0.73 g/cm3
D 0.88 g/cm
B
с

Answers

Answer:

Option C. 0.73 g/cm³

Explanation:

From the question given above, the following data were obtained:

Mass = 80 g

Area (A) = 10000 cm²

Thickness = 0.11 mm

Density =?

Next, we shall convert 0.11 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

0.11 mm = 0.11 mm × 1 cm / 10 mm

0.11 mm = 0.011 cm

Thus, 0.11 mm is equivalent to 0.011 cm.

Next, we shall determine the volume of the paper. This can be obtained as follow:

Area (A) = 10000 cm²

Thickness = 0.011 cm

Volume =?

Volume = Area × Thickness

Volume = 10000 × 0.011

Volume = 110 cm³

Finally, we shall determine the density of the paper. This can be obtained as follow:

Mass = 80 g

Volume = 110 cm³

Density =?

Density = mass / volume

Density = 80 / 110

Density = 0.73 g/cm³

Therefore the density of the paper is 0.73 g/cm³

Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.​

Answers

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI  

a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration​

Answers

Answer:

0

Explanation:

a = dv/dt

if v is constant than the slope of the v graph will be 0, so dv/dt is 0

a= 0

If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.

Answers

Answer:

17.8cm

Explanation:

1.3kg --> 4cm

1kg --> 3, 1/13cm

5.8kg --> 18.8cm

A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.

Answers

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i  the question;

mass of sun = M

minimum and maximum distance = r1 and r2 respectively

Now, using Kepler's third law,

" the square of period T of any planet is proportional to the cube of average distance "

T² ∝ R³

average distance a = ( r1 + r2 ) / 2

we know that

T² = 4π²a³ / GM

T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]

T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]

T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

T = 2π √[( r1 + r2 )³ / 8GM ]

Therefore, the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

An object is moving from north to south what is the direction of the force of friction of the object

Answers

Answer:

North

Explanation:

Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.

Answer:

South To North

Explanation:

Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north

a microwave operates at a frequency of 2400 MHZ. the height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. assume that microwave energy is generated uniformly on the uipper surface. What is the power output of the oven

Answers

Complete question is;

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly

downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

Answer:

Power ≈ 600,000 W

Explanation:

We are given;

Frequency; f = 2400 Hz

height of the oven cavity; h = 25 cm = 0.25 m

base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²

total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J

We want to find the power output and we know that formula for power is;

P = workdone/time taken

Formula for time here is;

t = h/c

Where c is speed of light = 3 × 10^(8) m/s

Thus;

t = 0.25/(3 × 10^(8))

t = 8.333 × 10^(-10) s

Thus;

Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))

Power ≈ 600,000 W

Physics help please

Answers

Answer:

i think the answer is 0.001m³

A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West

Answers

The start is located on theWest

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.

Answers

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, [tex]$\omega = \frac{1}{8}$[/tex]  rev/sec

                             [tex]$=\frac{2 \pi \times 7.5}{8}$[/tex]  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

[tex]$F=m.\omega^2.r$[/tex]

   [tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical

Answers

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

Other Questions
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