A 7.1 kg, 5.0 m board is balanced about a point 2.0 m from right end of the board. A cube is sitting centered 0.5 m from the right end. A 3.5 kg cylinder is dropped from a height of 2.0 m onto the left end of the board and sticks to it. The dropped cylinder is going 6.26 m/s just before it hits the very left end of the board. (Assume the cube and cylinder are point masses.)
a) What is the mass of the cube sitting 0.5 m from the right end?
b) What is the cylinder’s angular momentum about the board/cube's pivot point just before
hitting the board?
c) What is the total angular momentum of all of the masses just after the collision?
d) What is the total moment of inertia of the board and two masses just after collision?
e) What is the board's angular velocity just after collision?
f) How fast is the cube (originally 0.5 m from the right end) going just after the collision?

This is due tomorrow night, so take your time, but please make sure that it is done well, I want to understand how you got your answers.

Answer:

c- Time

Explanation:

The time will tell depending on the person who is exercising, the rate of time, the  workout time. & limits on weights....

Answer:

The guy above me is wrong the real answer is D

Explanation:

In order to calculate the amount of time needed for the boat to speed up to 70 m/s, we can use the formula (EQUATIONS):

Where V is the final speed, V0 is the initial speed, a is the acceleration and t is the time.

(GIVEN) So, for V = 70, V0 = 0 and a = 6, we can calculate the value of t (UNKNOWN):

(SETUP):

Therefore it will take 11.67 seconds (SOLVE).

The new acceleration of the object is therefore a' = a / 2.91 = 25.3 m/s/s / 2.91 = 8.69 m/s/s to 3 significant figures.

We know that the force exerted on an object is equal to its mass times its acceleration, or F = m * a. If the force exerted on the object remains constant, then the new acceleration of the object after its mass has been increased by a factor of 2.91 can be calculated as follows:

a' = F / (m * 2.91) = F / m'

Where m' is the new mass of the object.

Substituting the original acceleration and mass into the equation, we get:

a' = F / (m * 2.91) = F / (m * 2.91) = (m * a) / (m * 2.91) = a / 2.91

The new acceleration of the object is therefore a' = a / 2.91 = 25.3 m/s/s / 2.91 = 8.69 m/s/s to 3 significant figures.

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Answer:

What are the best flowers? (My opinion)

Explanation:

You're welcome.

Answer:

Tulips,

Lilies,

Lotuses on my opinion...

Answer:

8

Explanation:

Answer: Eight planets in our Solar System.

Explanation:

Answer:

30 . 7 m

Explanation:

It takes 2.5 seconds for the rock to fall the vertical height

d = 1/2 a t^2

d = 1/2 ( 9.81)(2.5^2) = 30.7 m

Answer:

simple carbohydrate

Explanation:

Simple carbs are like cakes, donuts, and candy, etc.

Hope this helps :D

If the magnification is -4 the the image is larger and is it inverted in nature.

The magnification of the body is given is -4, here the signs show the nature of the image and the numerical value is the times to which the object has been magnified. Here, it is inverted because there is negative sign and it is enlarged as the magnification is more than 1.

Therefore, a magnification of -4 indicates that the image is both inverted and larger than the object. The resultant picture will be a larger, inverted version of the object if it is seen via a lens or mirror with a magnification of -4.

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Given data:

*The cubic expansivity of brass is

The expression for the linear expansivity is given as

Substitute the known values in the above expression as

The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:

τ = r * F * sin(θ)

where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.

In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.

Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:

τ_M = r_M * F_M * sin(θ)

where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.

Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:

τ_M = τ_stick

r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)

Substituting the given values:

30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)

Solving for F_M:

F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)

F_M = 0.0264 kg * 9.8 m/s^2

F_M = 0.25872 N

Finally, we can convert the force into mass using the formula:

F = m * g

0.25872 N = M * 9.8 m/s^2

M = 0.0264 kg

Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

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Answer:

101.81818    MPH

Explanation:

Since no work is done, the power exerted is zero. Therefore, the man exerts no power on the wall.

In physics, force is defined as any action that can change the motion of an object or cause an object to accelerate. Force is a vector quantity, meaning that it has both magnitude (size or strength) and direction. The unit of force in the International System of Units (SI) is the Newton (N), which is defined as the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared (1 N = 1 kg × 1 m/s^2). Force can be measured using a variety of instruments, such as spring scales, strain gauges, or force plates. Some common types of forces include gravitational force, electromagnetic force, frictional force, and normal force. The study of forces and their effects on the motion of objects is known as mechanics and is a fundamental concept in physics.

Here,

a. The man does not do any work on the wall because the wall does not move. Work is only done when there is a displacement in the direction of the force applied.

b. Since no work is done, no energy is used or transferred.

c. The power exerted by the man can be calculated using the formula:

Power = Work / Time

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The correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.

Energy balance refers to the way in balance is achieved when intake of energy is equal to energy expended.

Energy refers to the impetus behind all motion and all activity. If is also the capacity to do work. Energy is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent.

So therefore, the correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.

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Answer: Hope this helps ;) don't forget to rate this answer !

Explanation:

There are two correct answers:

A) A corporation leader makes direct payment to the candidate.

D) A corporation contributes to a Super PAC that a PAC accepts contributions for a candidate.

Option A describes a scenario where a corporation directly donates money to a presidential candidate, which is allowed as long as it is done within the limits set by campaign finance laws.

Option D describes a scenario where a corporation donates money to a Super PAC, which is a type of political action committee that can accept unlimited donations from individuals, corporations, and other organizations. The Super PAC can then use the money to support or oppose a particular candidate, but it is not allowed to coordinate directly with the candidate or the candidate's campaign.

I hope this helps! Let me know if you have any other questions.

Answer:

The response to this question is as follows:

Explanation:

The whole question and answer can be identified in the file attached, please find it.

The force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.

The given parameters;

The force diagram of all the forces acting on the snowball is calculated as follows;

                                     ↑ N

                                     ⊕  → F

                                      ↓ W

Where;

Thus, the force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.

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The size of the universe is 41 orders of magnitude greater than that of the proton i.e. option A is correct.

The size of a proton is approximately 1 femtometer (fm), which is equivalent to 1 x 10^-15 meters. This is a very small size, much smaller than the size of an atom. And the size of the universe is in the order of 10^26 m. Comparing the size of both.

The size of a proton is 10^-15 m and the size of the visible universe is 10^26 m, so the difference in size can be calculated by dividing the size of the universe by the size of the proton :

log((size of universe)/(size of proton))

log(10^26/10^-15) = log(10^41) = 41

Therefore, option A stating  " the size of the universe is 41 orders of magnitude greater than that of the proton " is correct.

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Answer:

There are several modalities, or methods, of treatment: individual therapy, group therapy, couples therapy, and family therapy

Explanation:

Answer:

Modalities are therapeutic tools that we use to help our patients recover from surgery, injury or a condition that causes pain and dysfunction, so they can get back to the activities they enjoy.

Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.

Explanation:

The difference in the amplitude of the incident wave and the reflected wave is that the incident wave has a higher amplitude as compared to the reflected wave.

Both the incident and the reflected waves have fewer amplitudes as compared to the amplitude of the incident wave. If the tension in both strings is similar, then the wave speed is higher in the string with the lower mass density.

Velocity does not change due to reflection because the reflected wave has the same medium as the incident wave. The amplitude of the reflected wave can change due to reason some of the incident waves may be absorbed by the surface.

So we can conclude that the incident wave has a higher amplitude as compared to the reflected wave is the dissimilarity in the amplitude of the incident wave and the reflected wave.

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The current flowing in the 2Ω and 1Ω is 1.14 A and the current flowing in the 3Ω and 4Ω is 0.286 A.

The value of the current in each resistor is calculated by applying Kirchoff voltage law as follows;

The total voltage in loop 1 is calculated as;

2 + 4 - I₁R₁ - (I₁ - I₂)R₂ - I₁R₃ = 0

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

The current flowing in loop 2 is calculated as;

I = V/R

I₂ = ( 6 V - 4 V ) / (3 + 4)

I₂ = 0.286 A

The value of the current flowing in loop 1 is calculated as;

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

6 - 2I₁ - 3(I₁ - 0.286) - 1₁ = 0

6 - 3I₁ - 3₁ + 0.858 = 0

-6I₁ = -6.858

I₁ = 6.858 / 6

I₁ = 1.14 A

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Answer:

31.25m

Explanation:

using the formula

\(v {}^{2} = u {}^{2} + 2as\)

where v = final velocity = 0

u = initial velocity = 25

a = acceleration = -10 as its thrown upwards

s = distance (max height) = ?

rearranging the equation to find s:

\(s = \frac{ {v}^{2} - {u}^{2} }{2a} \)

\(s = \frac{0 {}^{2} - (25) {}^{2} }{2( - 10)} = 31.2m\)

Answer:

s the charge is positive, the electric field lines leave it at the point a d = 0.034 cm, the electric field lines go to the right.

Explanation:

The electric field around the cagases oisitvi, field leaving this means that the field lines move away from

If the charge is negative the lines of the field are directed towards the charge

Let's apply this to our case, as the charge is positive, the electric field lines leave it at the point a d = 0.034 cm, the electric field lines go to the right.

The cloud would begin to collapse if the temperature abruptly dropped as would the pressure, giving gravity the upper hand.

The pressure would rise with a rise in temperature, and the fog would start to grow.

The cloud would start to collapse if it gained a little mass because gravity would dominate.

In a molecular cloud, pressure, temperature, and gravity are all interconnected and play crucial roles in determining the cloud's properties and evolution.

Gravity is the force that holds the molecular cloud together and determines its overall shape and density. The more massive the cloud, the stronger its gravitational force and the tighter it can hold onto its gas and dust.

Temperature is related to the thermal energy of the gas and dust within the cloud. As the gas and dust particles move around, they collide with each other, transferring energy in the form of heat.

The pressure of a molecular cloud is determined by the temperature and density of the gas and dust within it. As the temperature increases, the pressure also increases. Similarly, as the density of the gas and dust increases, the pressure also increases.

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Complete question:

Suppose pressure and gravity are perfectly balanced within a certain molecular cloud. Describe what would happen to that balance if the temperature suddenly dropped. What would happen if the temperature suddenly rose? What would happen if the density increased without a change in temperature? What would happen if the cloud gained a little bit of mass?

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

yo did you ever get the answer?

Explanation:

Answer:

here ya go

Explanation:

When a force is applied to a car, the change in motion is proportional to the force divided by the mass of the car.

Answer:

Electric currents are created when electrons move from atom to atom within a solid metal wire. We measure currents in amperes (A). An electromagnetic field is created by connecting a piece of wire to both ends of a battery and wrapping the wire around a nail at the center. Turning on the electrical current produces a magnetic field.

Answer: Here you go, i hope this kinda helps.

Explanation:Disambiguation is just a fancy way of saying "asking clarifying questions".

Watson Assistant replies to user's questions based on a confidence score.

Sometimes the customer's question could be interpreted in two or three different ways.

For example, if you say you'd like to "book a table for 8", the assistant is able to ask a clarifying question:

Did you mean booking a table for 8PM, 8AM, or booking a table for 8 guests?

Watson Assistant will ask the question when its confidence score is divided between a few options to ensure that your customers get exactly the right service they need.