After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that the pendulum completes 98.0 full swing cycles in a time of 135s.

Required:
What is the value of g on this planet?

Answers

Answer 1

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

[tex]T = \frac{t}{n}[/tex]

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

[tex]T = \frac{135\ s}{98}[/tex]

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

[tex](1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}[/tex]

g = 11.2 m/s²


Related Questions

What is the difference between 1 celcius and 1 kelvin​

Answers

Answer:

One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point. The zero point on the Celcius scale was defined as the freezing point of water, which means that there are higher and lower temperatures around it.

A string with a length of 7.25 m is fixed at both ends. If the string is plucked so that there are five nodes along the string in addition to those at either end, what is the wavelength of the interfering waves for this mode?

Answers

Answer:

the  wavelength of the interfering waves for this mode is 2.4168

Explanation:

The computation of the wavelength of the interfering waves for this mode is shown below:

Given that

The length is 7.25m

And, the number of nodes is 6

So, the wavelength should be

= 2 × 7.25m ÷ 6

= 2.4168 m

Hence, the  wavelength of the interfering waves for this mode is 2.4168

The same would be considered

An astrophysicist mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 15.2 cm. They are separated by a distance of 40.2 cm. Lens 1 is to the left of Lens 2. In order to evaluate the lens combination as a single optical instrument, the teacher places an object 30.0 cm in front of (to the left of) Lens 1.

Required:
a. What is the final image's distance (in cm) from Lens 2? (Omit any sign that may result from your calculation.)
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?

Answers

Answer:

1 / i + 1 / o = 1 / f     thin lens equations

i = o f / (o - f)   rearranging

Lens 1:   object = 30 cm    f = 15.2 cm

i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm

o2 = 40.2 - 30/8 = 9.4 cm    distance of image 1 from lens 2

i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm

The final image is 24.6 cm to the left of lens 2

The first image is inverted

The second image is erect (as seen from the first image)

So the final image is inverted

M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69

It can be deduced that the final image's distance from Lens 2 will be 30.8 cm.

How to calculate the distance

By using the Lens formula, the distance will be calculated thus:

1/v + 1/u = 1/15.2

1/v + 1/30.0 = 1/15.2

v = 30.8cm

In this case, the image formed will be to the right.

Lastly, the overall magnification of the lens pair will be:

M = (30.8/30.0)[(-28.4/40.7 - 30.8)]

M = -2.94.

In conclusion, the magnification is -2.94.

Learn more about distance on:

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the electrostatic force between two objects is 40N. if the charge of one object is cut in half, and the distance is doubled, what is the new force?

Answers

Answer:

F1 = K Q1 Q2 / R1^2

F2 = K Q1 / 2 * Q2 / (2 R1)^2

F2 / F1 = 1/2 / 4 = 1/8

The new force is 5N   (1/2 due to charge and 1/4 due to distance)

Which of the following is NOT true about Potential Energy?
An object's position affects its potential energy
O Objects with more mass have more potential energy
Potential energy is Stored energy
Potential energy only occurs when an object is in motion

Answers

Answer:

Potential energy only occurs when an object is in motion.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Kinetic energy (K.E): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

b. Potential energy (P.E): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Additionally, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = P.E + K.E

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

Answers

Answer:

Wavelength = [tex]1.36\times 10^{-34}\ m[/tex]

Explanation:

Given that,

The mass of a ball, m = 0.113 kg

Velocity of the ball, v = 43 m/s

We need to find the wavelength of the ball. It can be calculated by applying the De-Broglie concept. So,

[tex]\lambda=\dfrac{h}{mv}[/tex]

Put all the values,

[tex]\lambda=\dfrac{6.63\times 10^{-34}}{0.113\times 43}\\\\=1.36\times 10^{-34}\ m[/tex]

So, the wavelength of the ball is equal to [tex]1.36\times 10^{-34}\ m[/tex].


3.9. How long will a boy sitting near the window of a train travelling at 36km/h see a train passing by in the opposite direction with a speed of 18km/h. The length of the slow-moving train is 90 m.​

Answers

Answer:

please give me brainlist and follow

Explanation:

Given:

The speed of the train ,which the boy is observing(t) = 18 km/hr

[ Now , to convert km/hr into m/s we have to multiply by 5/18 ]

So,

speed of the observed train = 18× 5/18

☞ 05 m/s

Also, The speed of the train in which the boy is sitting is(s) = 36 km/hr

Speed in m/s will be = 36 × 5/18

☞ 10 m/s

Also, The Length of the Train = 90m

Therefore:.

The speed of the train is given (expressed) by

= > speed \: = \frac{distance}{time}=>speed=

time

distance

= > time \: = \frac{distance}{Velocity }=>time=

Velocity

distance

Where,

velocity = Realtive velocity of both trains

Distance = Length of the train

So, The above mentioned formulae can also be written as :

= > time = \frac{length}{relative \: velocity}=>time=

relativevelocity

length

Here,

Relative velocity will be :

☞. s -(-t)

☞. 10 -(-5)

☞.15 m/s = Relative velocity

Now , Finally putting the values in the formulae respectively

■ Time = 90/15

■ Time = 6 sec [Answer]

s the gravitational force greater between the objects in Pair 1 or Pair 2? Explain why. (Picture shown below)

Answers

Answer:

Pair 1

Explanation:

The gravitational force greater between the objects in Pair 1 because in Pair 2 the objects are far apart and in pair 1 the object are more closer to each other.

So, Gravitational force directly proportional to distance

Thus, Increase in distance = Increased in gravitational force

-TheUnknownScientist

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

Answers

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

[tex] Wavelength = \frac {h}{mv} [/tex]

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]

Wavelength = 1.36 * 10^{-34} meters

The Earth Science students are making a human scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters (the length of a football field). How far from the school will the student representing Mars stand? A) 50 meters B) 105 meters 150 meters D) 1500 meters NEED HELP ASAP​

Answers

In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.

What is a unit of measurement?

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.

As given in the problem the  Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,

As given in the table marks is 1.5 AU far from Mars,

1 AU = 100 meters

1.5 AU = 1.5 × 100

           = 150 meters

Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.

Learn more about the unit of measurement here, refer to the link ;

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A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.70 ss apart. The speed of sound in air is 343 m/sm/s, and in concrete is 3000 m/sm/s. Part A How far away did the impact occur

Answers

Answer:

271.095 m

Explanation:

✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)

✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)

✓ Let the distance travelled by the sound = s

✓duration of Time that exist between heard of sounds = 0.70s

But we know that

Time = (Distance / Speed)

✓Time it takes the sound to travel through air= s/vi = s/343

✓Time it takes the sound to travel through concrete= s/vc = s/3000

✓ (s/343) - (s/3000) = 0.70

Finding LCM and simplify

[(3000s - 343s)]/1029000 = 0.70

2657s /1029000 = 0.70

Making " s" subject of the formula

s= (1029000 × 0.70)/2657

s=720300/ 2657

s= 271.095 m

Hence, The impact took place at a distance of 271.095 m away from the person.

A car has a mass of 1.00x10 to the 3rd power kilograms, it has an acceleration of 4.5 meters/seconds, what is the net force on the car?

Answers

Explanation:

Net force on the car= mass of the car × acceleration

F=1×10^3×4.5

=4.5×10^3 N

A person is dragging a packing crate of mass 74.9 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.35. He exerts a force F at and angle 43.0 degrees above the horizontal. What is the Force F such that the crate moves at a constant speed

Answers

Answer:

351.28 N

Explanation:

Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s².  So, f = μN = μW = μmg

So, the net force on the packing crate is

FcosФ - f = ma

FcosФ - μmg = ma

Since the crate moves at constant speed, its acceleration a = 0

So, FcosФ - μmg = ma

FcosФ - μmg = m(0)

FcosФ - μmg = 0

FcosФ = μmg

F = μmg/cosФ

Substituting the values of the variables into the equation, we have

F = μmg/cosФ

F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°

F = 256.907 kg-m/s²/0.73135

F = 351.28 kg-m/s²

F = 351.28 N

A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark
what is the value of x​

Answers

Answer:

X = 50 g

Explanation:

Please see attached photo for explanation.

From the attached photo,

Anticlock–wise moment = X × 20

Clockwise moment = 100 × 10

Anticlock–wise moment = clockwise moment

X × 20 = 100 × 10

X × 20 = 1000

Divide both side by 20

X = 1000 / 20

X = 50 g

Therefore, the value of X is 50 g

System A consists of a mass m attached to a spring with a force constant k; system B has a mass 2m attached to a spring with a force constant k; system C has a mass 3m attached to a spring with a force constant 6k; and system D has a mass m attached to a spring with a force constant 4k.
Rank these systems in order of increasing period of oscillation. (Use only the symbols < and =, for example A < B = C.)

Answers

Solution :

We know that the time period of oscillation of a spring mass system is given by :

[tex]$T = 2 \pi \sqrt{\frac{m}{k}}$[/tex]     , where m is mass and k is the spring constant

∴ [tex]$T_A = 2 \pi \sqrt{\frac{m}{k}}$[/tex]    .........(i)

  [tex]$T_B = 2 \pi \sqrt{\frac{2m}{k}}$[/tex] ..........(ii)

  [tex]$T_C = 2 \pi \sqrt{\frac{3m}{6k}} = 2 \pi \sqrt{\frac{m}{2k}}$[/tex]   ..........(iii)

  [tex]$T_D = 2 \pi \sqrt{\frac{m}{4k}}$[/tex]   ...............(iv)

Comparing the equations (i), (ii), (iii) and (iv)

We get

[tex]$T_B > T_A > T_C > T_D$[/tex]

So in increasing order of time period, we get

[tex]$T_D < T_C < T_A < T_B$[/tex]

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.70 ss for the boat to travel from its highest point to its lowest, a total distance of 0.660 mm . The fisherman sees that the wave crests are spaced a horizontal distance of 5.90 mm apart.

Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling ?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?

Answers

Answer:

a) 1.092 m/s

b) 0.33 m

c) 0.25 m

Explanation:

To start with, from the formula of wave, we know that

v = f λ, where

v = velocity of wave

f = frequency of the wave

λ = wavelength of the wave

Again, on another hand, we know that

T = 1/f, where T = period of the wave

From the question, we are given that

t = 2.7 s

d = 0.66 m

λ = 5.9 m

Period, T = 2 * t

Period, T = 2 * 2.7

Period, T = 5.4 s

If T = 1/f, then f = 1/T, thus

Frequency, f = 1/5.4

Frequency, f = 0.185 hz

Remember, v = f λ

v = 0.185 * 5.9

v = 1.092 m/s

Amplitude, A = d/2

Amplitude, A = 0.66/2

Amplitude, A = 0.33 m

If the other distance travelled by the boat is 0.5, then Amplitude is

A = 0.5/2

A = 0.25 m

PLEASE HELP

A ball, 1.8 kg, is attached to the end of a rope and spun in a horizontal circle above a student's head. As the student rotates the ball in a horizontal clockwise circle their lab partner counts 53 rotations in one minute. What is the ball's angular velocity in radians per second?

Answers

Answer:

The angular speed of the ball in radians per second is 5.55 rad/s.

Explanation:

Given;

mass of the ball, m = 1.8 kg

number of the ball's rotation per minute, n = 53 RPM

The angular speed of the ball in radians per second is calculated as follows;

[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]

Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.

un cubo de aluminio tiene un volumen de 45cm3 cuál es su masa en gramos

Answers

Material: aluminio volumen = v=64cm3 y masa = m

Please help 23 also 29 the answer choices are reflection or refraction your fraction absorption

Answers

Answer:

23. Option B. Hertz

29. Refraction.

Explanation:

23. Determination of the unit of measurement of frequency.

Frequency is simply defined as the number of oscillation made in one second. Mathematically, frequency can be represented as:

Frequency = 1/period

f = 1/T

Period is measured in seconds.

Thus, the unit of frequency becomes

f = 1/T

f = 1/s = s¯¹ (Hertz)

Therefore, the unit of frequency is Hertz.

29. When a wave enters a new medium, the speed of the wave is uttered. This leads to the bending of the wave. When this occurs, we can say refraction has taken place.

A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity 3.0m/s which statement below correctly describes the readings on the scales inside the elevator

Answers

Answer: b

Explanation:

The reading of the scale of the elevator  ascending with constant velocity is 150 N.

Reading of the scale

The reading of the scale on the elevaor is calculated by applying Newton's second law of motion;

R = m(a + g)

R = ma + mg

R = F + W

where;

a is the acceleration of the objects

At constant velocity, the acceleration of the object is zero (0).

R = 0 + 100 + 50

R = 150 N

Thus, the reading of the scale of the elevator  ascending with constant velocity is 150 N.

Learn more about reading of scale here: https://brainly.com/question/2516315

The energy of motion is called...?

Answers

The answer is Kinetic energy

The gravitational force between two objects is proportional to the square of the distance between the two objects.

a. True
b. False

Answers

Answer: True!

Explanation: The force is proportional to the square of the distance between 2 point masses

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 8.2. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Answers

Answer:

The correct answer is "3899.92 N".

Explanation:

The given values are:

Force,

[tex]F_{app}=58 N[/tex]

Ratio,

[tex]\frac{R_2}{R_1}=8.2[/tex]

As we know,

Area, [tex]A=\pi r^2[/tex]

or,

⇒  [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]

On substituting the value of "A", we get

⇒  [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]

⇒  [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]

On applying cross-multiplication, we get

⇒  [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]

On substituting the given values, we get

⇒       [tex]=58\times (8.2)^2[/tex]

⇒       [tex]=58\times 67.2[/tex]

⇒       [tex]=3899.92 \ N[/tex]

13. The percent of Earth's surface covered by high clouds
in January 1987 was closest to which of the following?
A. 13.09
B. 13.5%
C. 14.0%
D. 14.5%
16. Which of the following figures best represents the
monthly average cover of high. middle, and low clouds
in January 19922
E.
H.
cloud cover
okud cover
middle cloud
high clouds
high cloud
low clouds
low clouds
midle clouds
G.
J.
14. Based on Table I. a cosmic ray flux of 440.000 parti-
cles/m he would correspond to a cover of low clouds
that is closest to which of the following?
F. 28.75
G. 29.09
H. 29.35
J. 29.6%
ckud cover
cloud cover
high clouds
low clouds
middle cloud
high clouds
middle clouds
low clouds

Answers

B
Low clouds
H
High clouds

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:

Answers

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

Answer:

dorsiflexion

Explanation:

Hope this helps

The velocity of the source is positive if the source is ______________. Note that this equation may not use the sign convention you are accustomed to. Think about the physical situation before answering. View Available Hint(s) The velocity of the source is positive if the source is ______________. Note that this equation may not use the sign convention you are accustomed to. Think about the physical situation before answering. traveling in the x direction traveling toward the listener traveling away from the listener

Answers

Explanation is[tex]^{}[/tex] in a file

bit.[tex]^{}[/tex]ly/3tZxaCQ

A wave travels at a constant speed.How does the frequency change if the wavelength is reduced by a factor of 3 The frequency decreases by a factor of 3 the frequency increases by a factor of 3 the frequency increases by a factor of 9 the frequency does not change

Answers

Answer:

The frequency increases by a factor of 3.

Explanation:

The relation between speed, wavelength and frequency of a wave is given by :

[tex]v=f\lambda[/tex]

or

[tex]f\propto \dfrac{1}{\lambda}[/tex]

A wave travels at a constant speed. If the wavelength is reduced by a factor of 3, it would mean that the frequency increases by a factor of 3 because there is an inverse relationship between wavelength and frequency.

11) A tank of kerosene with density of 750 kg/m3 has a syphon used to remove the fluid that then exits into the local atmosphere, with pressure of 101 kPa. The pressure above the kerosene in the tank is 120 kPa absolute. The syphon tube has a diameter of 2 cm, exits the tank rising to 10 cm above the level of the kerosene and then drops down to 15 cm below the level of the kerosene where it exits into the atmospheric pressure. Calculate the exit velocity from the tube.

Answers

Answer:

[tex]7.32\ \text{m/s}[/tex]

Explanation:

[tex]v_1[/tex] = Velocity at initial point = 0

[tex]P_1[/tex] = Pressure in tank = 120 kPa

[tex]P_2[/tex] = Pressure at outlet = 101 kPa

[tex]\rho[/tex] = Density of kerosene = [tex]750\ \text{kg/m}^3[/tex]

[tex]Z_1[/tex] = Tank height = 15 cm

[tex]Z_2[/tex] = Height of pipe exit = 0

[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

From Bernoulli's equation we have

[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}[/tex]

The exit velocity from the tube is [tex]7.32\ \text{m/s}[/tex].

A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of mass 1.51 kg and length 1.79 meters spinning clockwise with an angular velocity of 5.12 rad/s is dropped on the spinning disk and stuck to it (the centers of the disk and the rod coincide). The combined system continues to spin with a common final angular velocity. Calculate the magnitude of the loss in rotational kinetic energy due to the collision

Answers

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

[tex]I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2[/tex]

The moment of inertia of the rod about its center is calculated as follows;

[tex]I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2[/tex]

The initial rotational kinetic energy of the disk and rod;

[tex]K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J[/tex]

The final rotational kinetic energy of the disk-rod system is calculated as follows;

[tex]K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J[/tex]

The loss in rotational kinetic energy due to the collision is calculated as follows;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J[/tex]

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

A honey bee's wings beat at 230 beats per second. If the speed of sound in air is 340 m/s, what is the wavelength of
the sound waves?

Answers

Answer:

[tex]from \: the \: wave \: equation \\ velocity = frequency \times wavelength \\ 340 = 230 \times \lambda \\ \lambda = \frac{340}{230} \\ \lambda = 1.5 \: m[/tex]

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