To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.

Answer:

 t = 14.53 s

Explanation:

The speed of a wave is constant and is given by

         v = \(\sqrt{ \frac{B}{ \rho} }\)

in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations

          v = x / t

           t = x / v

in this case the sound pulse leaves the ship and must return so the distance is

          x = 2d

where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s

let's calculate

           t = 2 10900/1500

           t = 14.53 s

The  wavelength at which the object emits the most radiation is 1.12 micrometers.

We use  Wien's displacement law in order to find the wavelength at which the object emits the maximum radiation.

The temperature of the item has an inverse relationship with the peak wavelength of the radiation it emits according to Wien's displacement law,

The formula of  Wien's displacement law,  is given as

λ_max = b/T

where λ_max =e peak wavelength of emitted radiation,

T = temperature of the object in Kelvin, and

b = Wien's displacement constant 2.898 × 10^-3 mK.

T = 2310°C + 273.15 = 2583.15 K

λ_max = (2.898 × 10^-3 mK) / 2583.15 K

λ_max = 1.12 × 10^-6 m = 1.12 μm

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Explanation:

some car crashes only produce minor injuries because the car crash isn't that bad.

The magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².

To find the magnitude of the angular acceleration of the roller, we can use the rotational analog of Newton's second law: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

First, let's calculate the moment of inertia of the roller. The moment of inertia of a solid cylinder rotating about its central axis is given by the formula: I = (1/2)mr², where m is the mass and r is the radius.

Given:

Mass of the roller (m) = 2.4 kg

Radius of the roller (r) = 3.8 cm = 0.038 m

Moment of inertia (I) = (1/2) * 2.4 kg * (0.038 m)² = 0.0021744 kg·m²

Next, we need to calculate the torque (τ) applied to the roller. Torque is given by the formula: τ = rFsin(θ), where r is the distance from the axis of rotation to the point of application of the force, F is the magnitude of the force, and θ is the angle between the force and the line connecting the axis of rotation and the point of application.

Given:

Force applied (F) = 16 N

Angle (θ) = 35°

Distance from the axis of rotation to the point of application (r) is equal to the radius of the roller, so r = 0.038 m.

Torque (τ) = (0.038 m) * (16 N) * sin(35°) = 0.2366 N·m

Now, we can use the equation τ = Iα and solve for the angular acceleration (α):

0.2366 N·m = (0.0021744 kg·m²) * α

α = 0.2366 N·m / 0.0021744 kg·m² ≈ 108.8 rad/s²

Therefore, the magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².

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Answer: Altostratus clouds they are gray or blue gray mid level clouds.

Explanation:

Answer:

F = 5.4 x 10⁻⁸ N

Explanation:

The gravitational force of attraction between two objects is given by Newton's Gravitational Law as follows:

F = Gm₁m₂/r²

where,

F = Gravitational Force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of student = 92 kg

m₂ = mass of pizza slice = 550 g = 0.55 kg

r = distance between student and pizza slice = 25 cm = 0.25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(92 kg)(0.55 kg)/(0.25 m)²

F = 5.4 x 10⁻⁸ N

Answer:

Explanation:

v² = u² + 2as

v² = 17² + 2(-9.8)(-51)

v² = 1,288.6

v = ±35.89707...

v = 36 m/s

Answer:

Clark Hull put the emphasis of his work on experimentation, an organized theory of learning, and the nature of habits, which he argued were associations between a stimulus and a response

Explanation:

Answer:

Ideas for Prototype Design  

Similar to a cardboard oven just smaller and solar powered

Preliminary Sketches (attach separate paper, if needed)  

Option A:  i gave the picture

____________________________________________________________

Advantages: Disadvantages:  

● the tin foil will attract the sun● if theres enough heat for the food to warm up

●the plastic will keep bugs and/or animal away from getting it ● it might take longer  

●the box gives the food something to be in instead of the ground  

Option B: i think im going to stay with my original plan i can always make changes

__________________________________________________________

More advantages and disadvantges

Advantages: Disadvantages:  

●reducing my carbon footprint by alot ● heat varies so the food my take really really long

●using no energy other than the sun  

__________________________________________________________

Which of the three designs will you move forward with? Explain your reasons for selecting this design.

I will be moving forward with option A bc i personally can't think of another way or design and i think option A has more advantages than disadvantages  

__________________________________________________________

Building the Prototype  

What modifications, if any, did you make to the basic design during the construction process?

I used skewers to hold the flap up

__________________________________________________________

Predictions  

Will your device warm the hot dog to a temperature of 165°F? How will your device do this?

Well i guess it depends n alot of things such as where i will place my solar cooker and the weather changes i will most likely put my solar cooker on the black paper because darker colors absorb more heat then lighter ones, then i will wait maybe 10 -20 minutes and check on my hotdog

__________________________________________________________

Will your device efficiently warm the hot dog in 20 to 25 minutes? How will your device do this?  so my device will do this by attracting the suns rays and energy towards the box

__________________________________________________________

Will your device reach a temperature well above that of its surroundings? How will your device do this?

I think so especially if i put it in direct sunlight around a darker surface and of course the aluminium foil will help with this as well

__________________________________________________________

Observations  

Record your observations and the results of the experimental tests of your device below. Temperature of the surroundings (in °F): ______68_______  

Maximum temperature reached inside the solar cooker (in °F): _____59_____

__________________________________________________________

i added a picture but i will also put the numbers here althought keep in mind u might want to switch them up because its based off of my temperatures

Use this table to record the internal temperature of the hot dog every 2 minutes. Use a separate sheet of  paper, if necessary.

time:                           Temperature (°F)

4:00                            20

4:02                            20

4:04                            25

4:06                            26

4:08                            30

4:10                            34

4:12                            38

4:14                            40

4:16                            44

4:18                            46

4:20                           50

4:22                           55

4:24                           57

4:26                           60

4:28                           65

4:33                           72

4:35                           75

4:37                           76

4:39                           80

__________________________________________________________

Evaluating Your Prototype  

What worked well?

i would say definitely the black paper and tin foil sin they attracted most of the suns energy

__________________________________________________________

Which features can be improved upon?

i think maybe if i did t on a different day it would have gotten hotter

__________________________________________________________

How could the overall design of this device be improved?  

Position the flap more so that the hotdogs can get more sun

Doing it on a hotter day

Letting it cook longer

__________________________________________________________

Why would this change be an improvement? What concepts related to thermal energy transfer is this  improvement based on?  

this change would be an improvement by the hotdogs getting warmed to the right temperature this is related to thermal energy by the position of the flap i think if i angled it more in an abtuse angle the sun would have transferred to the foil to the bottom foil with the hotdogs

__________________________________________________________

Sketch of Your Final Design  

Draw a well-labeled sketch of the final design.

i provided a sketch it should be the last picture

OKAY FINALLY PLZ DONT REORT THIS I WORKED SO HARD ON IT LAST TIME AND IT GOT DELETED IF U HAVE ANY QUESTIONS ABOUT THE PROJECT I WILL TRY TO HELP YOU IN THE COMMENTS THANK YOUUUU - ( if your wndering what grade i got on this got a 100 percent )

OH AND PEACE.

Answer:

I am simply adding another answer so the one above can get a brainliest

Explanation:

I do not care about the points that I will wind up getting for this and they are right here just answer with gibberish

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edit: in an attempt to return your points my question was taken down ill will return these

somehow

someway

sometime

Answer:

logistics is the process of storing, transporting and distributing goods to customers. The outbound logistics process starts with a customer sales order, moves on to warehouse packing and ends with product delivery.

Explanation:

Answer:

The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.

\(s\frac{d}{t}\)

Explanation:

hope this helps you my friend

The resultant of two displacement vectors having opposite directions is the difference of the two displacements, having the same direction as the smaller vector.

When two displacement vectors have opposite directions, it means they are pointing in opposite ways. In other words, one vector is in the opposite direction of the other. To find the resultant of these vectors, we need to subtract one vector from the other.

If we consider two displacement vectors, let's say vector A and vector B, and they have opposite directions, we can represent them as A and -B.

To find the resultant, we subtract vector B from vector A: A - (-B) or A + B.

The resultant will have the same direction as the smaller vector. This is because when we subtract a larger vector from a smaller vector, the resultant will have the direction of the smaller vector.

Therefore, the correct option is: "The resultant is the difference of the two displacements, having the same direction as the smaller vector."

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Answer:

I dont understand what you are trying to ask

Explanation:

Answer:

The characteristics that determine how easily two substances change temperature are:

The volume and density of the substances and the amount of time they are in contact do not directly affect how easily they change temperature.

Explanation:

Answer:

The dryer sheet is negatively charged and your hand is positively charged

Explanation:

Answer:

shivering raises your body temperature

Answer:

\(\huge\boxed{\sf a = 3.5 \ m/s^2}\)

Explanation:

Initial velocity = \(V_i\) = 0 m/s

Final velocity = \(V_f\) = 16.1 m.s

Time = t = 4.5 seconds

Acceleration = a = ?

\(\displaystyle a = \frac{V_f-V_i}{t}\)

Put the given data in the above formula.

\(\displaystyle a = \frac{16.1-0}{4.6} \\\\a = \frac{16.1}{4.6} \\\\a = 3.5 \ m/s^2\\\\\rule[225]{225}{2}\)

The amount the needle on the meter is deflected A. increases

This phenomenon can be explained by Faraday's law of electromagnetic induction. According to this law, when a magnetic field (created by the bar magnet) passes through a coil of wire, it induces an electric current in the wire. This induced current generates its own magnetic field, which interacts with the magnetic field of the bar magnet.

The deflection of the meter needle is a result of this induced current. When the number of coils of wire is increased, there is a greater number of wire loops for the magnetic field to pass through. This leads to a stronger induction of electric current, resulting in a larger deflection of the meter needle.

By increasing the number of coils, more magnetic flux is linked with the wire, resulting in a higher induced electromotive force (emf) and a greater current. This increased current produces a stronger magnetic field around the wire, leading to a larger deflection on the meter. Therefore, increasing the number of coils of wire enhances the magnetic field interaction, resulting in an increased deflection of the meter needle. Therefore, Option A is correct.

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The option that represents what the magnetic field look like above the North pole is an arrow that decreases as we go up and points up (E)

The magnetic field lines of a magnet point away from the north pole and towards the south pole. The field lines are strongest at the poles and weaken as you move away from the poles.

So, the arrow that represents the magnetic field above the north pole will be pointing up, but it will become smaller and smaller as you go up.

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Answer:

rolling friction

Explanation:

(a) The work done by the 150 N force is 877.5 Joules.

(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.

To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.

(a) The applied force's work may be estimated using the following formula:

Work = Force * Distance * cos(theta)

where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.

As a result, the applied force's work is:

Work = 150 N * 5.85 m * cos(0) = 877.5 J

So, the work done by the 150 N force is 877.5 Joules.

(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.

The frictional work may be estimated using the following formula:

Work = Force of Friction * Distance * cos(theta)

The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.

Thus, the work done by friction is:

Work = 150 N * 5.85 m * cos(180) = -877.5 J

Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.

The coefficient of kinetic friction may be calculated using the following equation:

Friction Force = Kinetic Friction Coefficient * Normal Force

The normal force equals the crate's weight, which may be computed as:

Normal Force = mass * gravity

where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N

Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:

Force of Friction = coefficient of kinetic friction * Normal Force

150 N = coefficient of kinetic friction * 343 N

coefficient of kinetic friction = 150 N / 343 N ≈ 0.437

As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.

In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.

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Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

As it hits the ground in time t, so put y = 0

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X =  16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

your welcome, and have a great day.

Answer:

a) 0.47MW

b) 39.24%

Explanation:

In order to find the power needed for the elevator to operate at its maximum capacity, we can make use of the following formula:

P=Fv

where P is the power, F is the force and v is the velocity.

The force the elevator must carry can be calculated with the following formula:

F=mg

where m is the mass of the elevator and g is the acceleration of gravity, so:

\(F=(8000 kg)(9.81 m/s^{2})\)

F=78 480 N

so now we can make use of the power formula:

P=Fv

P=(78 480N)(6 m/s)

P=470 880W

P=0.47W

b)

In order to find the efficiency, we will suppose that the generator can generate a maximum of 0.47 W so we use the following formula:

\(efficiency = \frac{P_{in}}{P_{out}}*100\%\)

\(efficiency=\frac{0.470880}{1.2}*100\%\)

efficiency=39.24%

The ratio of the heights from which A and B fell is 1:4.

The time of motion of a projectile is the time taken for a projectile to fall from the given height to ground level.

Mathematically, the formula for time of motion is given as;

t = √(2h/g)

h = ¹/₂gt²

where;

2h/t² = g

2h₁/t₁² = 2h₂/t₂²

h₁/t₁² = h₂/t₂²

let the time of motion of object A = t

let the time of motion of object B = 2t

height of fall of object A = h₁

height of fall of object B = h₂

h₁/h₂ = t₁²/t₂²

h₁/h₂ = (t²)/(2t)²

h₁/h₂ = (t²)/(4t²)

h₁/h₂ = 1/4

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The speed of the artillery shell when it hits its target is 100 m/s.

Given:

Initial vertical displacement (y) = 200 m

Vertical displacement at 100 m in the air (y') = 100 m

Final velocity in the vertical direction (vy') = 0 m/s (at the highest point of the trajectory)

Using the equation for vertical displacement in projectile motion:

y' = vy^2 / (2g),

where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can solve for the initial vertical velocity (vy).

100 m = vy^2 / (2 * 9.8 m/s^2),

vy^2 = 100 m * 2 * 9.8 m/s^2,

vy^2 = 1960 m^2/s^2,

vy = sqrt(1960) m/s,

vy ≈ 44.27 m/s.

Now, since the horizontal motion is independent of the vertical motion, the horizontal speed of the shell remains constant throughout its trajectory. Therefore, the speed of the shell when it hits its target is 100 m/s.

Hence, the speed of the artillery shell when it hits its target is 100 m/s.

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a) OA acceleration AB uniform speed travel BC acceleration CD uniform speed travel DE deceleration.

b). OA 80 AB 160 BC 45 CD 100 DE 50 d). 435 e). OA 80 AB 80 BC 40 CD 100 DE -200 87

acceleration: the rate at which the speed and direction of a moving object vary over time. When something moves faster or slower, it is considered to be accelerating. Motion on a circle accelerates even while the speed is constant because the direction is always changing.

According to given data;

A) OA acceleration AB moving at a constant speed BC acceleration CD traveling at a constant speed DE deceleration, according to the information provided.

B). OA 80

AB 80

BC 40

CD 100

DE -200

C.) OA 80

AB 160

BC 45

CD 100

DE 50

D). 435

E). 87

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Answer:75.3 seconds

Explanation:

Using conservation of momentum, the initial momentum of the cannon and the projectile is equal to the final momentum of the cannon and the projectile:

m_cannon * v_cannon = (m_cannon + m_projectile) * v_final

where m_cannon is the mass of the cannon, v_cannon is the initial velocity of the cannon, m_projectile is the mass of the projectile, and v_final is the final velocity of the cannon and projectile together.

Substituting the given values gives:

210 kg * 0 m/s - 1 kg * 610 m/s = (210 kg + 1 kg) * v_final

Solving for v_final gives:

v_final = -1.83 m/s

The negative sign indicates that the cannon and projectile are moving in the opposite direction after the firing.

The time required to bring the cannon to rest can be found using the equation:

F_resistive = m_cannon * a

where F_resistive is the resisting force, m_cannon is the mass of the cannon, and a is the acceleration of the cannon.

Substituting the given values gives:

1,700 N = 210 kg * a

Solving for a gives:

a = 8.10 m/s^2

The time required to bring the cannon to rest can be found using the kinematic equation:

v_final = v_initial + a * t

where v_initial is the initial velocity of the cannon, which is equal to the muzzle velocity of the projectile, and v_final is zero. Solving for t gives:

t = -v_initial / a

Substituting the given values gives:

t = -610 m/s / 8.10 m/s^2 = 75.3 s

Therefore, the time required to bring the cannon to rest is 75.3 seconds.

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