Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.

How do I proton and and electron compared

What’s the question?

A car starts from rest and accelerates for 7.2 s with an acceleration of 1.4 m/s2. How far does it travel? Answer in units of m.

**Answer:**

xn = 36.28 [m]

**Explanation:**

To solve this problem we must use the following equation of kinematics, which is ideal for a body that moves with constant acceleration.

[tex]x=x_{0}+(v_{o} *t)+(\frac{1}{2} )*a*t^{2}[/tex]

where:

x - xo = displacement of the car [m]

Vo = initial velocity = 0

t = time = 7.2 [s]

a = acceleration = 1.4 [m/s^2]

The initial velocity is zero, as the car begins its movement from rest.

xn = (x - xo), Now replacing

xn = (0*7.2) + 0.5*1.4*(7.2^2)

xn = 36.28 [m]

waht is science

wjwissbsskdldmndndnd

**Answer:**

the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

**Explanation:**

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.

A chef places an open sack of flour on a kitchen scale. The scale reading of

40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of

10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.

**Answer:**

**Explanation:**

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below

**Answer:**

40N

**Explanation:**

trust

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

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]

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

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:**

am I right? be honest

**Answer:**

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

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.

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!

Please provide explanation!!!

Thank you.

**Answer:**

(a) 102 cm/s

(b) 0.490 cm²

**Explanation:**

(a) Use Bernoulli equation.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0

½ ρ v₁² + ρgh₁ = ½ ρ v₂²

½ v₁² + gh₁ = ½ v₂²

½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²

v = 102 cm/s

(b) The flow rate is constant.

v₁ A₁ = v₂ A₂

(25.0 cm/s) (2.00 cm²) = (102 cm/s) A

A = 0.490 cm²

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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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.**

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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]

Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons

**Answer:**

**The electric potential energy of the system is 7.87x10⁻²⁰ J.**

**Explanation:**

The electric potential energy is given by:

[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]

Where:

q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C

r is the distance = 3x10⁻⁹ m

K: is the electrostatic constant = 9x10⁹ Nm²/C²

[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]

**Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.**

I hope it helps you!

The **electric potential energy** of the **system** should be **7.87x10⁻²⁰ J.**

SInce We know that

fdr = kq1q2/r

Here

q₁ = q₂ i.e. is the **charge** of the **protons** = 1.62x10⁻¹⁹ C

r should be the **distance **= 3x10⁻⁹ m

K should be the **electrostatic constant **= 9x10⁹ Nm²/C²

Now electric potential energy should be

= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) / 3x10⁻⁹ m

= 7.87x10⁻²⁰ J.

hence, The **electric potential energy** of the **system** should be **7.87x10⁻²⁰ J.**

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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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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

block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp

Answer:

Explanation:

For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.

The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction

Taking the sum of forces along the y component

Sum Fy = -W+R = ma

Since acceleration is zero

-W+R = m(0)

-W+R = 0

-W = -R

W = R

Hence the Normal reaction force acting on the on the body is equal to normal force

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.

Why is our (a person's) gravitational pull NOT as strong as the Earth's gravitational pull

on us?

Gravity doesn't act on us

The Earth is closer to us than the Moon

Our mass doesn't change so the pull is really the same

We are much smaller than the Earth.

**Answer:With gravity, two things with mass will want to move toward each other. However, we humans don't feel our gravity pulling on another person because it's not very big, but we do all feel the pull of Earth's gravity all the time - we're not all floating in the air, because that would be happening without Earth's gravity!**

**Explanation:**

A recipe gives the instructions below

After browning the meat pour off fat from the pan to further reduce fat use a strainer.

what type lf separation methods are described in the recipe

A decantation and screening

B distillation and screening

C decantation and centrifugation

D distillation and filtration

**Answer:**

**A. decantation and screening
**

**Explanation:**

Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.

**Answer:**

a

**Explanation:**

BRAINLIEST. Agraph is probelow. The graph shows the speed of a car traveling east over a 12 second period. Based on the information in the graph, it can be

that in the first second

Answer:speeding up constantly

Explanation:

The **graph **between the **time **and the **speed **of the car shows that the speed is **increasing **constantly, so, option C is **correct**.

A **moving **object's rate of change in **distance **traveled is measured as **speed**. Speed is a **scalar**, which implies it is a measurement with a magnitude but no **direction**.

A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.

**Given**:

The graph shows the speed of a car traveling east over a 12-second period,

As you can see from the graph, at time t = 0 sec the speed is 10 m/s,

At t = 3 sec, the speed = 15.3 m/s

At t = 6 sec, the speed = 20.3 m/s

Thus, speed is increasing constantly.

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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:**

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Converting compound units

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?

**Answer:**

Dividing the silicon density by 1000 and then multiply it by 1000000.

**Explanation:**

A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:

[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]

[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]

In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.

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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