To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize electric fields.

a. True
b. False

Answer:

Explanation:

Situation in which a mother holds most of the authority:

Matriarchy

Answer:

d. an object immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid.

Answer:

Gravity acts on all objects in the universe.

Explanation:

Answer:

The car's average acceleration points due west.

Explanation:

Resolving the acceleration will give a resultant due west

Answer:

Explanation:

Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.

PE = KE

mgh = ½mv²

    h = v²/2g

    h = 36.4²/ (2(9.81))

    h = 67.53109...

    h = 67.53 m

The boat must have moved, despite not being seen to move, because its relative position to the dock has changed. This phenomenon is known as relative motion .

Everything is always in motion, but the way we perceive it depends on our frame of reference.

In this scenario, the dock was the frame of reference for the initial position of the boat. When the boat moved to the cove, its position relative to the dock changed, and the dock was no longer an appropriate frame of reference. The boat's motion is now relative to the cove instead.

It is important to note that relative motion depends on the chosen frame of reference. If we were to choose the boat as the frame of reference, then it would be the dock that appears to move, not the boat. This is because motion is always relative to a chosen frame of reference.

In conclusion, the boat must have moved because its position relative to the dock changed. The concept of relative motion reminds us that motion is always relative to a chosen frame of reference, and that the way we perceive motion depends on our chosen frame of reference.

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In the experiment, place a ringing alarm clock inside a vacuum chamber, and observe that when the chamber is evacuated, the sound from the alarm clock cannot be heard, illustrating that sound does not travel through a vacuum.

To illustrate that sound requires a medium for its propagation and does not travel through a vacuum, you can perform the following experiment:

Take two identical bells or sound-producing devices and place them in separate chambers, one in a vacuum chamber and the other in a chamber filled with air.

Ensure that both chambers are isolated from external noise sources.

Activate the sound-producing devices simultaneously, creating sound waves in both chambers.

Observe and listen for the sound produced in each chamber.

In the chamber filled with air, you will hear the sound clearly, as air molecules transmit the sound waves, allowing them to propagate and reach our ears.

However, in the vacuum chamber, you will not hear any sound because there is no medium (such as air) for the sound waves to travel through. Without molecules to transmit the vibrations, the sound cannot reach our ears.

This experiment demonstrates that sound requires a medium, such as air or other substances, to propagate and be perceived by our ears. In a vacuum, where there is an absence of molecules, sound cannot travel.

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Answer:

What is the question though?

Explanation:

Answer:

37.4m/s

Explanation:

since the car doesn't accelerate, we can use the formula v=ωr where v is linear speed, ω is angular speed (rads/second) and r is radius. Substitute values for equation:

v=55*0.68

v=37.40

If a particle's velocity is nonzero, then its acceleration can be zero, because acceleration is the rate of change of velocity. If the velocity is constant and does not change, the acceleration is zero.

The rate of change of an object's velocity with respect to time is defined as acceleration. Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.

Because velocity is both a speed and a direction, there are only two ways to accelerate: modify your speed or your direction—or both.

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Option D. continental shelf

Continental shelf is a part of a continent that is submerged under an area of ​​relatively shallow water known as a shelf sea.

It is given in the question, that the shelf of undersea land reaches 200 meters and it also extends out from the shoreline. This condition is only possible when a structure is submerged, and such submerged region can only be called as continental shelf.

On the other hand guyots is a underwater volcanic mountain and deep-sea trenches are the the long narrow lowerings in the ocean floor with the depth upto 6000 meters. And abyssal plains are the underwater ocean floors which cannot be found at the depth of 200 meters.

Therefore continental shelf is used to described as a shelf of undersea land reaching a depth of about 200 meters (656 feet) and extending out from the shoreline.

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The fictional people and circumstances in realistic literature are meant to represent our society and reality. Growing up and facing personal and societal issues are the main topics.

In a word, the primary distinction between fiction and non-fiction is that fiction is the product of imagination, whereas non-fiction is entirely founded on facts and truth. To better grasp the two genres, let's now get in-depth with them.

While it may be based on a factual tale or circumstance, fiction is literature that was formed from the author's imagination and is not meant to be taken seriously as truth. Novels, short stories, and novellas are examples of literature in the fiction category.

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The mass of the chain in two decimal places is 2.55 kg.

The tension at the end of the chain is due to the weight of the weight supported by the chain.

The mass of the chain is calculated by applying Newton's second law of motion as follows;

W = mg

where

m is the mass of the chain

g is acceleration due to gravity

m = W/g

The mass of the chain is calculated as;

m = (25 N)/(9.8 m/s²) m = 2.55 kg

Hence, mass of the chain in two decimal places is 2.55 kg

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Answer:

If two bumper cars collide with a certain force, then they will move away from each other in opposite directions with the same force. This demonstrates Newton's third law, which states that for every action, there is an equal and opposite reaction.

Explanation:

We will determine the magnitude of the final component as follows:

So, the magnitude is approximately 8.36.

And the direction will be approximately 57.8° counter-clockwise.

This can be seeing as follows:

The amount of work done on the toboggan by the tension force of 105 N with a rope directed 32° above the snow is 3.12 KJ

W = F d cos θ

W = Work done

F = Force

d = Distance

θ = Angle between force and displacement vector

d = 35 m

F = 105 N

θ = 32°

W = 105 * 35 * cos 35°

W = 105 * 35 * 0.85

W = 3123.75 N m

W = 3.12 KJ

Work done is energy transferred to make an object move to a distance. Its unit is Joules which is denoted as J. It is the amount of work done by a force of 1 Newton to move a distance of 1 meter.

Therefore, the amount of work done on the toboggan by the tension force is 3.12 KJ

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Given that two large parallel plates are separated by 0.20 m and that the potential difference is 12V.

(a) Describe the motion of an electron released from rest at a distance "d" from the negative plate.

(b) What distance from the positive plate will the electron have a speed of 1 x 10^6 m/s?

For part (a):

The magnitude of an electric field can be given as  \(||\vec E||=\frac{\Delta V}{d}\), where "ΔV" is the potential difference and "d" is the distance between the plates.

So, \(||\vec E||=\frac{12 \ V}{0.20 \ m} \Longrightarrow \boxed{||\vec E||=60 \ \frac{N}{C} }\)

An electric field is created between the plates pointing from positive towards negative. We know that negative charges accelerate opposite the direction of electrical fields. So the electron placed "d" meters away from the negative plate will accelerate towards the positive plate at a constant rate.

For part (b):

We know that...

- the charge of an electron is \(\bold{-1.602 \times10^{-19} \ C}\).

- the mass of an electron is \(\bold{9.11 \times10^{-31} \ kg}\).

- \(\vec F_e=q\vec E\)

- \(\vec F =m\vec a\)

\(\Longrightarrow \vec F_e=(-1.602 \times10^{-19} \ C)(60 \ \frac{N}{C} }) \Longrightarrow \boxed{\vec F_e= -9.612 \times10^{-18} \ N}\)

\(\Longrightarrow \vec F =m\vec a \Longrightarrow \vec a=\frac{\vec F}{m} \Longrightarrow \vec a=\frac{-9.612 \times10^{-18}}{9.11 \times10^{-31} \ kg} \Longrightarrow \boxed{\vec a=-1.06 \times10^{13} \ m/s^2}\)

Kinematic Equation: \(\vec v_f^2=\vec v_0^2+2\vec a \Delta \vec x\)

\(\Longrightarrow 1 \times10^{12} \ m^2/s^2=-2.11 \times10^{13} \ m/s^2 \Delta \vec x \Longrightarrow \Delta \vec x= \frac{1 \times10^{12} \ m^2/s^2}{-2.12 1\times10^{13} \ m/s^2}\)

\(\Longrightarrow \boxed{\Delta \vec x= -0.047 \ m}\)

The distance from the positive plate we'll call, "D."

\(D=0.20+\Delta \vec x\)

\(\Longrightarrow D=0.20+\Delta \vec x \Longrightarrow D=0.20 \ m+(-0.047 \ m) \Longrightarrow \boxed{D=0.153 \ m} \therefore Sol.\)

The volume of an ice block ​will be 0.543 m³.

Density is defined as the mass per unit volume. It is an important parameter to understand the fluid and its properties. Its unit is kg/m³.

The mass and density relation is given as

mass = density × volume

The given data in the problem is as ;

v is the volume of ice block =?

ρ is the density = 920kg/m³

m is the mass of ice block = 500kg

The mass and density relation is given as;

mass = density × volume

500 kg =920kg/m³ ×v

v=0.543 m³

Hence, the volume of an ice block ​will be 0.543 m³.

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One parameter of the system that is held constant that can be considered is the mass of the water.

In science, the constant is a type of unmodified variable that remains equal along with the experimental procedure.

The constants of a particular system must be considered and they cannot change in experiments or observations.

In conclusion, one parameter of the system that is held constant that can be considered is the mass of the water.

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The electric field at the y-axis position is -10.0 N/C and is oriented towards the negative charge and electric potential at the point on the y-axis is -0.5 V..

To calculate the electric potential and electric field at the point on the y-axis, use equations for electric potential and electric field due to an electric dipole:

Electric potential V = kq/(r_+) - kq/(r_-)

Electric field E = kq/(r_+)² - kq/(r_-)²

where k = Coulomb constant,

q = charge of each point charge,

r_+ = distance from the positive charge to the point on the y-axis, and

r_- = distance from the negative charge to the same point on the y-axis.

Using the given values:

r_+ = √((0.01 m)² + (0.1 m)²) = 0.1005 m

r_- = √((0.008 m)² + (0.1 m)²) = 0.1002 m

q = 3.0 μC

Electric potential:

V = (9.0 x 10⁹ N·m²/C²)(3.0 x 10⁻⁶ C)/(0.1005 m) - (9.0 x 10⁹ N·m²/C²)(3.0 x 10⁻⁶ C)/(0.1002 m)

V = 215.6 V - 216.1 V

V = -0.5 V

The electric potential at the point on the y-axis is -0.5 V.

Electric field:

E = (9.0 x 10⁹ N·m²/C²)(3.0 x 10⁻⁶ C)/(0.1005 m)² - (9.0 x 10⁹ N·m²/C²)(3.0 x 10⁶ C)/(0.1002 m)²

E = 2705.6 N/C - 2715.6 N/C

E = -10.0 N/C

The electric field at the point on the y-axis is -10.0 N/C, directed towards the negative charge.

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Answer:

a= v-v÷t

= 9-6÷1.5

= 3÷1.5

a = 2m/s²

(1) The acceleration of the car is determined as 17.42 m/s².

(2) The acceleration of the crate is determined as 8.89 m/s².

The acceleration of the car is calculated from the net force acting on the car.

∑F = ma

F - Ff = ma

F - μW = ma

where;

m = W/g

m = (9320)/(9.8)

m = 951.02 kg

2(9320) - 0.222(9320) = 951.02a

16,570.96 = 951.02a

a = 17.42 m/s²

F - μW = ma

F - μmg = ma

302 - 0.75(18.6 x 9.8) = 18.6a

165.29 = 18.6a

a = 8.89 m/s²

Thus, the acceleration of the car is determined as 17.42 m/s².

The acceleration of the crate is determined as 8.89 m/s².

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Answer:

The vertical component of the velocity can be found using the formula:

V₀y = V₀ * sin(θ)

where V₀ is the initial velocity, θ is the angle of projection, and V₀y is the vertical component of the velocity.

Substituting the given values, we have:

V₀y = 48 * sin(53°)

Using a calculator, we can evaluate sin(53°) to be approximately 0.799:

V₀y = 48 * 0.799

V₀y ≈ 38.352

Therefore, the vertical component of the velocity with which the ball was launched is approximately 38 meters/second, which corresponds to option B.

Answer:

B.) 38 meters/second

Explanation:

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:

c h i c k e n n u g g e t s . . . . . .

jk its 2.6

The tension in cable B is 87.5 N, see the computation in the section below

Tension in cable A = 800sin30 = 400 N

Tension in cable B = 800cos30 = 693.33 N

Total Tension = 400 + 693.33 = 1093.33 N

Tension in cable B = 1093.33 - 800 = 293.33 N

Tension in cable B = 293.33/2 = 87.5 N

The tension in cable B, attached to an adjoining building, is 87.5 N in order to hang a sign weighing 800 N so that cable A, attached to the store, makes a 30.0° angle.

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