**Answer:**

*B. ***12**

**Explanation:**

*4 x 3 = ***12**

The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm

**Answer:**

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

**Explanation:**

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]

[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] **(Eq. 1)**

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.

[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] **(Eq. 1b)**

Where:

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.

[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.

If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:

[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]

[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster

Answer: The options are not given.

Here are the options.

a) There is an additional force lifting up on you.

(b) At the top you continue going straight and the seat moves out from under you.

(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (

d) Both b and c are correct.

(e) a, b, and c are correct.

The correct option Is D.

B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.

Explanation:

At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.

That is the scenarion that occur...

Examine the diagram below.

Which of the above objects is moving the fastest?

A.

All 3 objects are moving at the same speed.

B. B

C. C

D. A

**Answer:**

*Correct choice: D. Object A is the fastest*

**Explanation:**

In a distance vs time graph, the distance is plotted vertically, and the time is plotted horizontally.

The diagram shows three graphs of objects A, B, and C.

The graph of A shows the object traveled 12 meters in 3 seconds, for a speed of 12/3= 4 m/s.

The graph of B shows the object traveled 8 meters in 4 seconds for a speed of 8/4=2 m/s.

Finally, the object C travels 4 meters in 4 seconds, for a speed of 4/4= 1 m/s

Thus, **the fastest object is A.**

plzzz helppp

You are pushing a box North in the hallway, at 20n, and a friend gets in front of the box and goes in the opposite direction, at 30n. What direction is the box going at? How much force does the box have going in that direction?

**Answer:**

the box is going south at 10n

**Explanation:**

A designer is creating an obstacle for an obstacle course where a person starts on a moveable platform of height H from the ground. The person grabs a rope of length L and swings downward. At the instant the rope is vertical, the person lets go of the rope and attempts to reach the far side of a water-filled moat. The left side of the moat is directly below the position where the person will let go of the rope. The designer runs several tests in which the rope has different lengths and moves the platform such that the rope is always initially horizontal. The designer notices that the person cannot land on the other side if the length L is very short. The designer also notices that the person also cannot land on the other side if the length L is very close to the height H.

Assume the size of the person is much smaller than the lengths L and H. Let D represent the horizontal distance from below the release point to where the person lands.

Required:

a. Why does the person land in the moat if the rope's length is very short?

b. Why does the person land in the moat if the length is nearly the same as the height of the platform?

**Answer:**

* when L → H chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

the speed of the platform is very small, so we do not have the minimum required value

vox = √ (g / (2 (H)) D

**Explanation:**

For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.

Let's start with the projectile launch

as the body leaves the vertical its velocity must be horizontal

x = v₀ₓ t

y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero

t = √ 2 y₀ / g

we substitute

x = vox √2y₀ / g

v₀ₓ = √(g / 2y₀) x

In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)

v₀ₓ = √(g /(2 (H -L)) D

this is the minimum speed to cross the well.

Now let's use conservation of energy

starting point. On the platform

[tex]Em_{o}[/tex] = U = m g H

final point. At the bottom of the swing

Em_{f} = K + U = 1 / 2m v² + m g (H -L)

as there is no friction the mechanical energy is conserved

Em_{o} = Em_{f}

m g H = 1 / 2m v² + m g (H -L)

v = √ (2gL)

let's write our two equations

the minimum speed to cross the well

v₀ₓ = √ (g /(2 (H -L)) D

the speed at the bottom of the oscillatory motion

v = √ (2g L)

we analyze the extreme cases

* when L → H chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

the speed of the platform is very small, so we do not have the minimum required value

vox = √ (g / (2 (H)) D

From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well

V₀ₓ = v

g / (2 (H -L) D² = 2g L

4 L (H- L) = D²

4 H L - 4 L2 - D² = 0

L² - H L - D² / 4 = 0

let's solve the quadratic equation

L = [H ± √ (H2-D2)] / 2

we assume that H> D

L = ½ H [1 + - RA (1 - (D / H) 2)]

The two values of La give the range of values for which the two speeds are equal

A) The **person lands** in the moat if the **rope's length** is very short because :

B) The **person lands** in the moat if the rope length is similar to the **height** of the platform because :

Following the assumptions;

size of the person is much smaller than L and H

D = horizontal distance

The conditions that will cause the person to land on the moatTheHence we can conclude that The **person lands** in the moat if the **rope's length** is very short because The **speed** of the platform is less than the required **minimum speed** and The **person lands** in the moat if the rope length is similar to the **height** of the platform because,the** speed** required to cross the moat approaches **infinity.**

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Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

**Answer:**

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

**Explanation:**

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

[tex]v = r\cdot \omega[/tex] **(Eq. 1)**

Where:

[tex]r[/tex] - Radius of rotation of the particle, measured in meters.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

[tex]v[/tex] - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

[tex]\omega = \frac{\theta}{t}[/tex] **(Eq. 2)**

Where:

[tex]\theta[/tex] - Angular displacement, measured in radians.

[tex]t[/tex] - Time, measured in seconds.

By applying **(Eq. 2)** in **(Eq. 1)** we get that:

[tex]v = \frac{r\cdot \theta}{t}[/tex] **(Eq. 3)**

From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:

[tex]s = r\cdot \theta[/tex]

[tex]v = \frac{s}{t}[/tex]

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field

**Answer:**

The maximum torque on the loop is **395.80 N.m.**

**Explanation:**

Given;

number of turns of the wire, N = 150 turns

length of the square loop, L = 18.0 cm = 0.18 m

current in the wire, I = 50.9 A

Magnetic field, B = 1.6 T

Maximum torque on the loop is given by;

τ = NIAB

τ = (150)(50.9)(0.18²)(1.6)

τ = **395.80 N.m**

Therefore, the maximum torque on the loop is **395.80 N.m.**

It took 50 joules to push a chair 5 meters across the floor. With what force was the chair pushed?

**Answer:**

The chair was pushed with 10 N.

**Explanation:**

The chair was pushed with 50 Joules.

Work = Force * Distance

50 J = F * 5m

F = 50 / 5 = 10N

The chair was pushed with 10 N.

The chair was pushed with **10 N **force.

**Work **is defined as the measure of **energy transfer **that occurs when an object is moved over a distance by an external force, at least part of which is applied in the direction of **displacement**.

If the force is **constant **then **work **can be calculated by multiplying the length of the path by the component of the force acting along the path, which is expressed mathematically as work W equal to the force f over a distance d, or **W = fd. **

So, for above given information,

Work done= 50 joules

Distance covered by the chair = 5m

Then, Force= W/d

=50/5= 10N

Thus, the chair was pushed with **10 N **force.

Learn more about **Work done**, here:

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HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the

motion of the ball?

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

A tower crane has a hoist motor rated at 159 hp. If the crane is limited to using 72.0 % of its maximum hoisting power for safety reasons, what is the shortest time in which the crane can lift a 5550 kg load over a distance of 89.0 m

Answer:

The value is [tex]t = 56.68 \ s [/tex]

Explanation:

From the question we are told that

The rating of the hoist motor is [tex]k = 159hp = 159 *746 =118614 \ W[/tex]

The percentage of it power used is [tex]z = 0.72 * 118614=85402.08 \ W[/tex]

The mass of the load is m = 5550 kg

The distance is h = 89.0 m

The potential energy required to lift the load through that distance is

[tex]E = m * g * h[/tex]

=> [tex]E = 5550 * 9.8 * 89.0[/tex]

=> [tex]E = 4840710 \ J[/tex]

Generally the time taken is mathematically represented as

[tex]t = \frac{E}{ z}[/tex]

=> [tex]t = \frac{4840710}{ 85402.08}[/tex]

=> [tex]t = 56.68 \ s [/tex]

Magnets are usually made up of which material

A. plastic

B. iron ore

C. copper

D. gold

**Answer:**

B. iron ore

**Explanation:**

Hope this helps

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If 0.5 kg of this material is used in a transformer core, how long would it have to operate at a frequency of 60 cps to heat up 1

**Answer:**

Hello your question is incomplete attached below is the complete question

answer : 49 seconds

**Explanation:**

considering only Hysteresis loss

we have to calculate the Area affected/under the Hysteresis loss

= volume * area

= 4 * ( 1.5 * 20 ) = 120 tesla. A/m

**next we calculate the volume of the material**

= mass of material / density

= 500 grams / 7.9 g/cm^3 = 6.33 * 10^-5 m^3

**next we calculate the heat lost per cycle **

= 6.33 * 10^-5 m^3 * 120 = 0.007596 joules

**The total heat required to raise temperature by 1°c**

= Cp * ΔT * n

= 3R * n * ΔT = 3(8.314) * 8.95 * 1 = 223.23 Joules

where n = number of moles = 500grams / 55.85 = 8.95moles

ΔT = 1

Therefore the time required to have to operate at a frequency of 60 cps

= Total heat required / heat lost per cycle

=( 223.23 / 0.007596 ) 60 cps

= 489.796 seconds ≈ 49 seconds

Two particles are separated by 0.38 m and have charges of -6.25x 10 C and 2.91 x 10 C. Use Coulomb's law to predict the force between the particles if the distance is doubled. The equation for Coulomb's law is Fe = g, and the constant, k, equals 9.00 x 10° Nm/C A. -1.13 x 10-6 N OB. 1.13x 106N O C. 2.83 x 10-7 N OD.-2.83x 10N sUBMIT

**Answer:**

I do not understand what you are asking

A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.

Answer:

2.096m/s

Explanation:

The speed of this soccer ball can be calculated using the formula;

Speed = distance/time

According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;

Speed = 6.5/3.1

Speed of the ball = 2.096m/s

Therefore, the speed of the ball before hitting the tree is 2.096m/s

Silly Goose falls 1.0 m to the floor. How long does the fall take

**Answer:**You need to give more explanation sorry

**Explanation:**

**Answer:**

4.20 seconds

**Explanation:**

Supposing that silly goose weighs 69 pounds, we need to start on the math.

Simple maths, truly and really. 69/1=69, of course.

Therefore it will take 4.20 seconds for silly goose to hit the ground. if he is going to be a silly goose though, he can just go in the pond, instead of wasting his time.

Research has shown that this type of interview is the most effective in predicting later job

performance.

**Answer:**

Situational Interview

**Explanation:**

A situational interview is about as close to the real job as it gets. During this type of interview, candidates may be presented with a visual or audio simulation of a scenario and asked to respond to it. They are asked to analyze a problem and profer suggestions on how they would handle it.

If the candidate has solved similar problems in the past, it will come to the fore.

If they haven't then the best outcome is that it will tell the interviewers how well the candidate is able to solve similar problems.

An example of a Situational Interview question is this:

**How would you handle an angry customer who for no justifiable reason has decided to create a problematic scene to disrupt the business?**

Because Situational Interviews are about behavioral responses (present, past, and future), they are powerful tools in determining or predicting future job performance. An interviewing technique that is developed using this methodology is called the S.T.A.R.

This is an acronym for **Situation, Task, Action, Result.**

Situation: the candidate is asked to present a challenging situation that occurred recently. *This tests what the candidate sees as a challenging situation.*

Task: The candidate based on the situation is asked to identify what they need to do to remedy the problem. *This tells the interviewer(s) whether or not the candidate can think up a solution for the problem.*

Action: Here they define the actual steps taken to resolve the problem

Result: The candidate against the above is required to give the result gotten

**Action** and **Result **tell the interviewer the quality of the candidate's ability to follow through and achieve the intended results. This also judges the quality of execution in terms of cost and time. The candidate with the lowest cost and time and the highest quality of outcome is considered the best.

Cheers

when hydrogen shares electrons with oxygen the outermost shell of the hydrogen atoms are full with how many electrons? and oxygens valence shell is full with how many electrons? because the valence shells of these atoms are full,the atoms are stable.

**Answer:**

2 and 8

**Explanation:**

please mark me brainiest I would really appreciate it.

How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)

A) 100 J B) 200 J C) 300 J D) 400 J

**Answer:**

B) 200 [J]

**Explanation:**

In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]

F = m*g

where:

m = mass = 5 [kg]

g = gravity acceleration = 10 [m/s^2]

F = 5*10 = 50 [N]

w = F*d

where:

F = force = 50 [N]

d = 4 [m]

w = 50*4 = 200 [J]

One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.

Answer:

The value is [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]

Explanation:

From the question we are told that

The speed of the airplane from the north is [tex]\frac{dN}{dt} = -181 \ km /hr[/tex]

The negative sign is because the direction is towards the south

The speed of the airplane from the east is [tex]\frac{dE}{dt} = -278 \ km/hr[/tex]

The negative sign is because the direction is towards the west

The distance of the southbound plane from the airport is [tex]N = 30 \ km[/tex]

The distance of the westbound plane is [tex]E = 15 \ km[/tex]

Generally the distance between the plane is mathematically represented using Pythagoras theorem as

[tex]R^2 = N^2 + E^2[/tex]

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

[tex]2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}[/tex]

Here

[tex]R = \sqrt{N^2 + E^2}[/tex]

=> [tex]R = \sqrt{30^2 + 15^2}[/tex]

=> [tex]R = \sqrt{30^2 + 15^2}[/tex]

=> [tex]R =33.54 \ m [/tex]

[tex]2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)[/tex]

=> [tex] 67.08 * \frac{dR}{dt} = -19200[/tex]

=> [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]

The **rate **of change of the **distance **between the **planes **is **286.23 km/hr.**

The given parameters;

The **distance **between the two **planes **is calculated by applying **Pythagoras theorem** as shown below;

[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]

The **rate **of **change **of the **distance **between the **planes **is calculated as follows;

[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]

Thus, the **rate **of change of the **distance **between the **planes **is **286.23 km/hr.**

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Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion is reversible but not isothermal and that the temperature decreases as the expansion proceeds. (a) Find an expression

**Answer:**

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

**Explanation:**

what is the meaning of the word physics

**Answer:**

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

**Explanation:**

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Grass and plants get energy from

А

the sun.

B

eating food.

с

windmills.

D

electrons.

A is the answer for that question

**Answer:**

From the Sun

**Explanation:**

Plants can't eat any food. They don't ue or need windmills to get energy. They are plants so they don't have any electrons. The only way that they can recive energy from is the sun. Sometimes plants die when they don't get enough sun because they don't have any energy to live.

An object is rolled at 12 m/s down a table. It stops

after 15s. What was its acceleration?

Variables:

Equation and Solve:

**Answer:**

**We are given:**

initial velocity (u) = 12 m/s

final velocity (v) = 0 m/s

time taken (t) = 15 seconds

acceleration (a) = a m/s²

**Solving for acceleration:**

from the first equation of motion

**v = u + at**

replacing the variables

0 = 12 + (a)(15)

0 = 15a + 12

a = -12 / 15

**a = -4 / 5 m/s²**

am I right? be honest

**Answer:**

I chose c because it is the greater slope at point c

A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?

The force 120 Newton’s

A .05 kg rubber ball is dropped and hits the floor with an initial velocity of 10 m/s. It rebounds away from the floor with a final speed of 7 m/s after being in contact with the floor for .01 seconds. Find the magnitude of the force exerted by the floor on the rubber ball.

**Answer:the answer is 3**

**Explanation:**

1. What is Ohm"s law?

2. If you placed a negatively charged hairbrush near your hair, what charge would your hair be?

3. You must change a lightbulb and the new lightbulb has a larger resistance. If the voltage of the battery does not change, what happens to the current going through the flashlight?

HELLPPPP

1. Ohm's law shows the relationship between:

voltagecurrentresistanceFormula: voltage = current x resistance

2. The negative charge on the hairbrush will induce a positive charge on your hair. As a result, your hair is going to be attracted to the hairbrush (and repelled by other strands of hair.)

3. V = IR, so if the resistance of the current increases, and the voltage of the current stays the same, there is as a result, going to be less current.

Best of Regards!

A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement

**Explanation:**

Let east = E, and, west = opposite to east = - E.

Here, displacement:

=> 2m east + 4m west + 1m east

=> 2E + 4(-E) + 1E

=> 2E - 4E + 1E

=> - 1E

=> 1(-E)

=> 1m west

And, distance,

=> 2m + 4m + 1m = 7m

The **distance **of a person is **7 m** and the **displacement **of the person is **1m west.**

To **find **the distance and displacement, the given values are,

A person **walks **2.00 m east, then turns and goes 4.00 m west, then turns and **goes **back 1.00 m east.

**Displacement**:

**Distance**:

Let us **consider **East = E and west = **opposite **to east = - E.

**Calculating **the **displacement**:

= 2m east + 4m west + 1m east

= 2E + 4(-E) + 1E

= 2E - 4E + 1E

= - 1E

= 1(-E)

= 1m west.

The **displacement **is **1m west**.

Now **calculating **the distance,

= 2m + 4m + 1m

= 7m

The **distance **is **7m.**

Thus, the displacement and the distance is found as **1 m west and 7m.**

Learn more about **distance **and **displacement**,

https://brainly.com/question/3243551

**#SPJ6**

True.or false A railroad track runs southwest to northeast.

**Answer:**

ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]

Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.

As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]

Contents

1 Definitions in American context

2 History

2.1 Faster inter-city trains: 1920–1941

2.2 Post-war period: 1945–1960

2.3 First attempts: 1960–1992

2.4 Renewed interest: 1993–2008

2.5 Plans for 2008–2013

3 Current state and regional efforts

3.1 The Northeast

3.1.1 Northeast Corridor: Next Generation High-Speed Rail

3.1.1.1 Proposed routes

3.1.2 Northeast Maglev proposal

3.1.3 New Jersey–New York City upgrades

3.1.4 New York

3.1.5 Pennsylvania

3.2 Western States

3.2.1 California

3.2.2 Pacific Northwest

3.2.3 Arizona

3.3 Mid-Atlantic and the South

3.3.1 Florida

3.3.2 Southeast

3.3.3 Texas

3.4 Midwest

3.4.1 Illinois and the Midwest

3.5 The Southwest

4 Federal high-speed rail initiatives

4.1 American Recovery and Reinvestment Act of 2009

4.1.1 Strategic plan

4.2 2009 federal grant funding

4.3 2010 allocation

4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida

4.4 2011 and 2012 proposals and rejections of funding

5 See also

6 Notes

7 Further reading

8 External links

**Explanation:**

A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person

**Answer:**

20000

**Explanation:**

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000

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