An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?

Answers

Answer 1

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

[tex]\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%[/tex]

so, the efficiency of heat engine is 13.7%.


Related Questions

A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet

Answers

Answer:

it’s c not d

Explanation:

took the test

Answer: D!!!

Explanation: jus got it wrong from the other answer.

Objects accelerate because

Answers

Friction I think soooooooooo

Thorium^+2

Chemical symbol:
Atomic Number:
Mass: 232
# of protons
# of neutrons
Group #
Period #

Answers

Answer:

chemical symbol: Th

atomic number:90

protrons :90

neutrons:142

group#:4

period#: 9

Explanation:

you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons

The scientific term describing the ball changing position as it goes from
Steph Curry's hand, into the air, and through the hoop.
A= velocity
B= speed
C= vector
D= motion

Answers

Answer:

Vector

Explanation:

Vector is a quantity that shows the direction or path through which a body travels with as it changes position.

As body travels, the direction sometimes changes and this is described by the vector of the body.

Velocity is a vector quantity that describes the displacement per unit of time of a body.

Speed is a scalar that deals with the distance covered per time

So, a vector specifies the magnitude of a physical quantity and also the direction through which it travels.

As a bicycle is ridden west in a straight line with decreasing speed,the acceleration of the bicycle must be

Answers

Answer:

Decreasing

Hope this helps! :)

A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?

A 300N
B 202N
C 398N
D 98N

Answers

Answer:98

Explanation:hope this helps!

A 5.0 kg block is pushed 2.0 m at a con-
stant velocity up a vertical wall by a constant
force applied at an angle of 30.0° with the
horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s2.
F
30°
2 m
5 kg
Drawing not to scale.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.

Answers

Answer:

[tex]W_F=127.64283 J[/tex]

Explanation:

Information Given:

[tex]m = 5kg[/tex]  [tex]v=constant[/tex]

Key: μ = Kinetic Friction (Kf)   θ = Theta α = 180° N = Normal Force

[tex]W_F=F_ydcos[/tex]θ

[tex]W_F=Fdsin[/tex]θ

[tex]_{net}F_y = sin[/tex]θ-μ[tex]N-mg=0[/tex]

[tex]_{net}F_x = 0[/tex]

[tex]N=Fcos[/tex]θ

[tex]Fsin[/tex]θ-μ[tex]N=mg[/tex]

[tex]Fsin[/tex]θ-μ[tex]Fcos[/tex]θ[tex]=mg[/tex]

[tex]F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex] →[tex]W_F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex]

[tex]W_F=\frac{(2)(9.81)(2)sin(30)}{sin(30)-(0.40)cos(30)}[/tex]

[tex]W_F=127.64283 J[/tex]

Which change in an object would increase the force needed to move the object?
А.
decreasing the velocity of an object
B
increasing the volume of an object
с
decreasing the mass of an object
D
increasing the mass of an object

Answers

Answer:

D i think

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop.

Answer:

Increasing the mass of the object (option D in the list of answers)

Explanation:

Recall that F = m x a

therefore, if the mass increases, the force increases

What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm.

Answers

Complete Question

What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. The student has 70-cm-long arms

Answer:

The value is  [tex]w__{rpm} } = 29.17 \ rpm[/tex]

Explanation:

From the question we are told

    The distance from the handle to the bottom of the bucket is  [tex]d = 35 \ cm = 0.35 \ m[/tex]

      The length of the students arm is  L = 70 cm  = 0.70  m

   Generally the acceleration due to gravity experienced by the bucket of  water is mathematically represented as

       [tex]g = w^2 * r[/tex]

Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as

       [tex]r = L + d[/tex]

So

         [tex]g = w^2 * ( L + d )[/tex]

= >     [tex]w = \sqrt{\frac{g }{ L + d } }[/tex]

= >     [tex]w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }[/tex]

= >     [tex]w = 3.055 \ rad/s[/tex]

Generally the angular speed in revolution per minute is mathematically represented as

        [tex]w__{rpm} } = \frac{w * 60 }{2 \pi }[/tex]

=>      [tex]w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }[/tex]

=>      [tex]w__{rpm} } = 29.17 \ rpm[/tex]

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?

Answers

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

A diffraction grating with 68 slits per cm is used to measure the wavelengths emitted by hydrogen gas.
A. At what angles in the fourth-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm?
B. What are the angles if the grating has 12,800 slits per cm?

Answers

Answer:

a

  [tex]\theta _1 =0.687 ^o[/tex]

  [tex]\theta _2 =0.630 ^o[/tex]

b

 Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.1 will not be valid

Explanation:

From the question we are told that

     The slit grating is  [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]

      The order of spectrum is [tex]n = 4[/tex]

Generally the width of the slit is mathematically represented as  

               [tex]a = \frac{1}{ 6800}[/tex]      

=>            [tex]a = 0.000147 \ m[/tex]  

Generally the condition for constructive interference is

              [tex]asin\theta = n * \lambda[/tex]

Now for the first wavelength the angle is evaluated as

             [tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]

    =>     [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]

    =>    [tex]\theta _1 =0.687 ^o[/tex]

Now for the second wavelength the angle is evaluated as

             [tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]

    =>     [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]

    =>    [tex]\theta _2 =0.630 ^o[/tex]

Gnerally if grating is   [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]

Generally the width of the slit is mathematically represented as  

               [tex]a = \frac{1}{ 1280000}[/tex]      

=>            [tex]a = 7.813 *10^{-7} \ m[/tex]  

Generally the condition for constructive interference is

              [tex]asin\theta = n * \lambda[/tex]

Now for the first wavelength the angle is evaluated as

             [tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]

            [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]

    =>     [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

    =>    [tex]\theta _1 =0.687 ^o[/tex]

Now for the second wavelength the angle is evaluated as

             [tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]

    =>     [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]

     =>  [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N, 180.0 on it. If the coefficient of friction is 0.18, calculate the deceleration rate of the wagon as it is caught.

Answers

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

[tex]\sum F_x = ma_x\\[/tex]

[tex]Fm - Ff = ma_x\\[/tex] where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

[tex]Ff = \mu R\\Ff = \mu mg\\[/tex]

[tex]\mu[/tex] is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

[tex]Fm - Ff = ma_x\\Fm - \mu mg = ma\\[/tex]

Substitute the given parameters

[tex]Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2\\[/tex]

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

A statement of the second law of thermodynamics is that:__________.
a) spontaneous reactions are always exothermic.
b) energy is conserved in a chemical reaction that has a decrease in entropy.
c) spontaneous reactions are always endothermic.
d) in a spontaneous process, the entropy of the universe increases.

Answers

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of  randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."  

The universe here refers to the system's disorder and the disorder of the surroundings.  Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

What is the algebraic sign (+ or -) for designating the distance and velocity vectors for an object in free fall?

Answers

Answer:

positive- downward

negative- upward

Explanation:

A vector is a quantity described by a magnitude and a direction.

The algebraic sign, positive (+) and negative (-) are used to designate the distance and velocity vectors for an object in free fall where positive represents downward  and negative represents an upward movement of the object.

Hence, the correct answer is "positive- downward  and negative- upward".

The change in motion (acceleration) of an object depends on
The size of the force
The mass of the object
BOTH the size of the force AND the mass of the object

Answers

Answer:

BOTH the size of the force AND the mass of the object

Explanation:

Acceleration of an object is the rate of change of its velocity.

The relation between force, mass and acceleration is given by the formula as follows :

F = ma

m is mass

a is acceleration

It would mean that the change in motion or the acceleration of an object depends on both the size of the force and the mass of the object. Hence, the correct option is (c).

A 60-kg jogger runs up a long flight of
stairs in 4.0 s. The vertical height of the
stairs is 4.5 m.
a. Estimate the jogger's power
output in watts and horsepower.
b. How much energy did this
require?

Answers

Explanation:

Given parameters:

Mass of Jogger  = 60kg

Time  = 4s

Vertical height  = 4.5m

Unknown:

Jogger's power output  = ?

Energy required  = ?

Solution:

The power output of the jogger is defined as the rate at which work is done.

 Power  = [tex]\frac{force x distance}{time}[/tex]  

Now insert the parameters and solve;

  Work done = Force x distance = mgh

   m is the mass

   g is the acceleration due to gravity = 9.8m/s²

    h is the height

  Work done  = 60 x 9.8 x 4.5  = 2646J

 Power = [tex]\frac{2646}{4}[/tex]   = 661.5W

Energy required;

 The work done here is also the energy required;

 Energy required  = 2646J

The energy required is 2646 J and the jogger's power output 661.5 W.

Given here,

Mass of Jogger  = 60 kg

Time  = 4s

Vertical height  = 4.5 m

The power output of the jogger is defined as the rate of energy tranfer  work is done.

Power  =  W/t

Where,

W- work = mgh = 60 kg  x  9.8 m/s² x 4.5 m = 2646 J

t - time - 4 s

Put the values, we get

Power = 661.5W

To know more about power,

https://brainly.com/question/8288959

A ball is thrown off a cliff at a speed of 10 m/s in a horizontally direction. The ball reaches the ground 1.5 seconds. If the ball is launched a second time at the same speed from a second higher cliff, which of the following is true?
A. The ball takes a longer time and lands further away from the foot of the cliff.
B. The ball takes longer to hit the ground, but lands at the same distance from the foot of the cliff.
C. The ball takes the same time lands at the same distance from the foot of the cliff.
D. The ball falls further away from the foot of the cliff, but takes the same time.

Answers

i am pretty sure it is d

When liquid water gets into cracks of rock and freezes, it __ and ___.

Answers

It expands and pushes the crack further aprt

An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life?

Answers

Answer:

The half-life is [tex] t_{1/2} = 1.005 h[/tex]

Explanation:

Using the decay equation we have:

[tex]A=A_{0}e^{-\lambda t}[/tex]

Where:

λ is the decay constantA(0) the initial activityA is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]

[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]

[tex]0.5=e^{-\lambda*1 h}[/tex]

Taking the natural logarithm on each side we have:

[tex]ln(0.5)=-\lambda[/tex]

[tex]\lambda=0.69 h^{-1}[/tex]

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]

[tex] t_{1/2} = 1.005 h[/tex]

I hope it helps you!

if chilled coke and hot tea are
kept together tea cools down but ko gets warm why

Answers

Tea gets cool beacuse of heat tranfer through convenction
Heat transfers from a area of hot region to cold.
The tea is hot in this case, due to taht factor the heat will move towards teh cooler region which is coke and will make the coke warm


Hope this helped!

What would its weight be on Jupiter?
24.9N

Answers

The weight would be 62.92.

Answer:

1.898 × 10^27 kg

Explanation:

thats how much it ways

A monatomic ideal gas with an initial pressure of 500 kPa and an initial volume of 1.80 L expands isothermally to a final volume of 5.20 L. How much work is done on the gas in this process?
A) 1700J
B) 875J
C) 1570J
D) 900J
E) 955J

Answers

Answer:

955 J  

Explanation:

PV = nRT

500 x 10³ x 1.8 x 10⁻³ = nRT

= 900 J

work done by gas in isothermal expansion

= nRT lnV₂ / V₁

= 900 ln 5.2 / 1.8

= 900 x ln 2.89

= 900 x 1.06

= 955 J  

A 500 kg car is moving at 30 m/s. The driver sees a barrier ahead. If the car takes 100 m to come to rest, what is the magnitude of the force necessary to stop the car?

How do you solve this question?

Answers

Answer:

F = 2250 [N]

Explanation:

In order to solve this problem, we must first use the following equation of kinematics.

[tex]v_{f}^{2} =v_{o}^{2}-2*a*x[/tex]

where:

Vf = final velocity = 0 (come to rest)

Vo =  initial velocity = 30 [m/s]

a = acceleration or desaceleration [m/s²]

x = distance = 100 [m]

[tex](0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}][/tex]

Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force [N]

m = mass = 500 [kg]

a = acceleration = 4.5 [m/s²]

[tex]F = 500*4.5\\F = 2250 [N][/tex]

. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it by 10.0 cm?

Answers

Ans    4 more to be exact

Explanation:

The force applied to the spring will be "400 N".

Given values are:

Spring constant,

k = 40 N/m

and,

x = 10 m

The force will be:

→ [tex]F = kx[/tex]

By substituting the values, we get

      [tex]= 40\times 10[/tex]

      [tex]= 400 \ N[/tex]

Thus the above answer is right.

Learn more about spring constant here:

https://brainly.com/question/15277652

Forces of 70 N at 130 degrees, and 20 N at an angle of 280 degrees, measured counter-clockwise from the positive x-axis, act on an object.
A. What are the components (F1x, F1y) of the first force force (in Newtons)?
B. What are the components (F2x, F2y) of the second force force (in Newtons)?
C. What are the components (Fx, Fy) of the resultant force (in Newtons)?
D. What is the magnitude of the resultant force (in Newtons)?
E. What is the angle of the resultant force with respect to x-axis?

Answers

Answer:

A. ) F₁ₓ = -45.0 N F₁y = 53.6 N

B.)  F₂ₓ = 3.48 N F₂y = -19.7 N

C.)  Fₓ = -41.5 N Fy = 33.9 N

D)  F = 53.6 N

E)  θ = -39. 2º (320.8º)

Explanation:

A)

Applying simple trig, like definitions of cos and sin of an angle, we can get the x- and y- components of F₁, as follows:

       [tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]

B)

Repeating for F₂:

       [tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]

C)  

The x- and y- components of the resultant force, are just the algebraic

        sum of the x- and - y components of F₁ and F₂:

Fₓ = Fₓ₁ + Fₓ₂ = -45 N + 3.48 N =  -41.5 N (5)By the same token, Fy can be written as follows:Fy = Fy₁ + Fy₂ = 53.6 N + (-19.7 N) = 33.9 N (6)

D)

The magnitude of the resultant force can be obtained applying the Pythagorean Theorem to Fx and Fy, as follows:

       [tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]

E)

Finally the angle regarding the x- axis of the resultant force vector, can be obtained using the definition of the tangent of an angle, as follows:

       [tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]

Under state law, what is the blood-alcohol limit for legally operating a motor vehicle?

Answers

Answer:

HOPE IT HELPS.....

Explanation:

The legal standard for drunkenness across the United States ranges from 0.10 to 0.08. If a person's BAC measures 0.08, it means that there are 0.08 grams (i.e., 80 mg) of alcohol per 100 ml of blood. The American Medical Association says that a person can become impaired when the blood alcohol level hits 0.05.

Answer: 0.08

Driving While Intoxicated or Impaired is Illegal

Under Michigan law, it is illegal to drive: While intoxicated, or impaired, by alcohol, controlled substance, or other intoxicating substance. With a bodily alcohol content of 0.08 or more. (This crime is one of the driving while intoxicated offenses.)

Hope this helps... Stay safe and have a Merry Christmas!!!!!!!! :D

John decided to cycle to his friend's house at a speed of 5km/h and the journey took 2
hours. How far did John cycle?

Answers

Answer:100 miles

Explanation:

Answer:

10 Km

Explanation:

He is going 5km per hour and he arrived at his friend's house in 2 hours. You multiply 5 by 2 and you get 10.

n
Question 4
1 pts
A bus travels on an interstate highway at an average speed of 90 km/hrs. How far does it take to travel
in 30 mins? The distance equals speed times time, or d = st.
O 45 km
O 98 Km
O 56 km
O 432 Km

Answers

[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]

A hare can run at a rate of 15 m/s, while a turbocharged tortoise can now crawl at a rate of 3 m/s, how much of a head-start (time-wise) does the tortoise need in order to tie the hare in a 250 meter race?

A.
16.7 seconds

B.
66.7 seconds

C.
83.3 seconds

D.
100 seconds

Answers

Answer:

t = 66.7 s

Explanation:

Given that,

Speed of a hare, v = 15 m/s

Speed of a turbocharged tortoise, v' = 3 m/s

The hare in a 250 meter race

Let the Hare takes time t. It can be calculated as follows :

[tex]t=\dfrac{250}{15}=16.67\ s[/tex]

Let a turbocharged tortoise takes t'. It can be calulated as follows :

[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]

To tie the race, required time is given by :

[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]

Hence, the correct option is (b) i.e. 66.7 seconds.

One of the harmonics of a column of air in a tube that is open at both ends has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column in this tube?

Answers

Answer:

The fundamental frequency is  [tex]f_1 =128 \ Hz[/tex]

Explanation:

From the question we are told that

   The frequency of one harmonics is  [tex]f_x= 448 \ Hz[/tex]

    The next higher harmonic is  [tex]f_z = 576 \ Hz[/tex]

Generally the frequency of an air column open at both ends is mathematically represented as

              [tex]f_n = \frac{nv }{ 2 L }[/tex]

Here n  is the order of the harmonics (frequency)

        v is the velocity of the sound

        L  is the length of the column

So for one harmonics we have that

        [tex]f_k = \frac{n v }{2L}[/tex]

Then for the next higher harmonics

       [tex]f_x = \frac{n+1 ) v}{2 L }[/tex]

Generally the difference between these frequencies is mathematically represented as  

       [tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]

=>    [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]

=>    [tex]\frac{ v }{2L} = 128[/tex]

Generally for fundamental  frequency n =  1

So  

       [tex]f_1 = n * \frac{v}{2L}[/tex]

So

       [tex]f_1 =1 * 128[/tex]

=>    [tex]f_1 =128 \ Hz[/tex]

Other Questions
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