Which change(s) of state require an increase in energy?

Answer:

F-ma

Explanation:

If you are speaking of objects like satellites, etc. then their mass is much less than that of the Earth. A good approximation is Newton's first law of motion:

Force = Mass ×  Acceleration

often written:

F = m a

The gravitational force is the same between the Earth and the object - only the mass differs. So the acceleration is inversely proportional to the mass.

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = [tex]q_{int}[/tex] /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

Answer:

Net force = 11340 N

Explanation:

Given that,

Mass of an automobile, m = 2100 kg

Acceleration of the automobile, a = 5.4 m/s²

We need to find the net force required for the automobile. The net force is the product of mass and acceleration. It can be given by the formula as follows :

[tex]F=ma\\\\F=2100\ kg\times 5.4\ m/s^2\\\\F=11340\ N[/tex]

So, the net force is 11340 N.

Answer:

28 m/s

Explanation:

vf^2=vi^2+2a(delta y)

=sqrt(2* -9.8 m/s^2* -40 m)

=28 m/s

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Answer: A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes.

Answer:

C a position from which something is observed

om edu 2021

Explanation:

Explanation:

s=d/t

t= d/s

t= 2/40

t= 0.05 hr

Answer:

The force exerted by the rope is FT = 225.06 [N]

Explanation:

In order to solve this problem we must use a static analysis, since Globe does not move. For a better understanding in solving this problem, a free body diagram with the forces acting on the globe is attached.

The buoyant force acts upward as it causes the balloon to tend to float, the weight of the balloon tends to lower the balloon and the downward tension force does not allow the balloon to float

The buoyant force is defined by the following equation:

FB = Ro*V*g

where:

FB = Buoyant force [N]

Ro = density of the air = 1.3 [kg/m^3]

V = volume of the balloon = 20 [m^3]

g = gravity acceleration = 9.81 [m/s^2]

FB = 1.3*20*9.81 = 255.06 [N]

Now we do a sum of forces equal to zero in the y-axis

FB - 30 - FT = 0

255.06 -30 = FT

FT = 225.06 [N]

Answer:

[tex]Velocity=\frac{dx}{dt}[/tex]

Explanation:

Remember that instantaneous velocity is just a measure to know the velocity that any object has at any point given in time, so we just need to know the distance it has travel, which would be the change in position, and the time it took that change in position to occurr, this means distance by time, so we just divide dx by dt and we have the solution for instantaneous velocity.

If a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex] change in time, the instantaneous velocity will be given by,

[tex]v = \dfrac {dx}{dt}[/tex]

[tex]v = \dfrac {dx}{dt}[/tex]

Where,

[tex]v[/tex] - instantaneous velocity

[tex]dt[/tex]   - change in distance (position)

[tex]dt[/tex]- change in time

Therefore, if a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex]change in time, the instantaneous velocity will be given by,

[tex]v = \dfrac {dx}{dt}[/tex]

Learn more about  Instantaneous velocity.

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Answer:

The average kinetic energy

The amount of heat a substance has or the average kinetic energy of particles in a substance

Answer:

1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor

Answer:

The answer is Top-Down processing

Explanation:

I had this question on a apex quiz and i got it correct.

Answer:

a) [tex]\beta = 1.111\times 10^{-7}[/tex], b) [tex]\beta = 9\times 10^{-7}[/tex], c) [tex]\beta = 3.087\times 10^{-6}[/tex], d) [tex]\beta = 2.5\times 10^{-5}[/tex], e) [tex]\beta = 0.5[/tex], f) [tex]\beta = 0.877[/tex]

Explanation:

From relativist physics we know that [tex]c[/tex] is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) A car traveling 120 kilometers per hour:

At first we convert the car speed into meters per second:

[tex]v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)[/tex]

[tex]v = 33.333\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 33.333\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 1.111\times 10^{-7}[/tex]

b) A commercial jet airliner traveling 270 meters per second:

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 270\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 9\times 10^{-7}[/tex]

c) A supersonic airplane traveling Mach 2.7:

At first we get the speed of the supersonic airplane from Mach's formula:

[tex]v = Ma\cdot v_{s}[/tex]

Where:

[tex]Ma[/tex] - Mach number, dimensionless.

[tex]v_{s}[/tex] - Speed of sound in air, measured in meters per second.

If we know that [tex]Ma = 2.7[/tex] and [tex]v_{s} = 343\,\frac{m}{s}[/tex], then the speed of the supersonic airplane is:

[tex]v = 2.7\cdot \left(343\,\frac{m}{s} \right)[/tex]

[tex]v = 926.1\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 926.1\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 3.087\times 10^{-6}[/tex]

d) The space shuttle, travelling 27000 kilometers per hour:

At first we convert the space shuttle speed into meters per second:

[tex]v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)[/tex]

[tex]v = 7500\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 7500\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 2.5\times 10^{-5}[/tex]

e) An electron traveling 30 centimeters in 2 nanoseconds:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

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

Where:

[tex]v[/tex] - Speed, measured in meters per second.

[tex]d[/tex] - Travelled distance, measured in meters.

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

If we know that [tex]d = 0.3\,m[/tex] and [tex]t = 2\times 10^{-9}\,s[/tex], then speed of the electron is:

[tex]v = \frac{0.3\,m}{2\times 10^{-9}\,s}[/tex]

[tex]v = 1.50\times 10^{8}\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 1.5\times 10^{8}\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 0.5[/tex]

f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

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

Where:

[tex]v[/tex] - Speed, measured in meters per second.

[tex]d[/tex] - Travelled distance, measured in meters.

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

If we know that [tex]d = 10^{-14}\,m[/tex] and [tex]t = 0.38\times 10^{-22}\,s[/tex], then speed of the electron is:

[tex]v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}[/tex]

[tex]v = 2.632\times 10^{8}\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 2.632\times 10^{8}\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 0.877[/tex]

Answer:

maybe the answer is in is D part

Answer:

Explanation:

Answer:

The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.

Explanation:

To convert from gram/centimeter³ to kilogram/meter³

First,

NOTE:

1 kilogram = 1000 gram, that is 1 gram = 0.001 kilogram; and

1 meter = 100 centimeter, that is 1 centimeter = 0.01 meter

then,

1 meter³ = 1000000 centimeter³, that is

1 centimeter³ = 0.000001 meter³

Now, we can write that

1 kilogram/meter³ = 1000 gram / 1000000 centimeter³

= 0.001 gram/centimeter³

If 1 kilogram/meter³ = 0.001 gram/centimeter³

Then, 1 gram/centimeter³ = 1/0.001 kilogram/meter³  = 1000 kilogram/meter³

Hence, 1 gram/centimeter³  = 1000 kilogram/meter³

∴The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.

That is,

2.33 gram/centimeter³ will be 2.33 × 1000 kilogram/meter³

= 2330 kilogram/meter³

Answer:

shortening its length.

The period of a pendulum is not dependent on the mass of the ball, but only on the length of its string. The period of a pendulum may be decreased by shortening the length of pendulum

The shortening of the pendulum length often affect its period.  The length of the string do influence the pendulum's period making it to be like the longer the length of the string, the longer the pendulum's period.

A pendulum that has a longer string often has a lower frequency. That is, it will swings back and forth in little times in a given amount of time than when the pendulum is with a shorter string length.

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Answer:

They are perpendicular.

Explanation:

Answer:

Negative

Explanation:

duh

Answer:

:)

Explanation:

negative.

go add the snap carmel.bratz

Answer:

Making an observation

Explanation:

The use of senses as a tool to gather information is described as making an observation.

While making an observation, the senses must be at alert.

Therefore, the act of using senses or tools to gather information is called making an observation.

Answer:

Did a little research.

Explanation:

The Law of Action-Reaction:

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum). According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club collides with a golf ball at rest upon a tee, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations. Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects that causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.

Consider the collision between a moving seven ball and an eight ball that is at rest in the sport of table pool. When the seven ball collides with the eight ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven ball experiences a leftward force that causes it to slow down; the eight ball experiences a rightward force that causes it to speed up. Since the two balls have equal masses, they will also experience equal accelerations. In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Answer:

D) 12 sodium, 4 phosphorus. 16 oxygen

Explanation:

there is a four in front of the full formula, so you multiply all of the sub numbers by 4

Answer:

36m/s

Explanation:

v=u+at

v=21+(2.2×6.9)

v=21+15.18

v=36.18

v=36m/s

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

Hope this helps.

Answer:

1176 Nm or J

Explanation:

W = F*d

F = 60kg * 9.8 kgm/s^2 = 588 N

W = 588 N * 2m = 1176 N*m

The formula for calculating work is Work = Force Distance. The SI unit for work is the joule (J) or Newton meter (N m). One joule is the amount of work done when one Newton of force moves a one-meter object.

1176 N × m = work

When energy is transferred from one store to another, work is performed. When a force causes an object to move, work is done.

W = F.d

F = 60kg × 9.8 kgm/s²= 588 N

W = 588 N × 2m = 1176 N × m.

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Answer:

I NEED OPTIONS

Explanation:

Answer:

v = - 1,715 m / s ,   x = 0.0156 m

Explanation:

This is an oscillatory movement exercise, which is described by the expression

          x = A cos (wt + Ф)

we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero

As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration

         v = dx / dt

         v = - A w sin wt

          a = - Aw² cos wt

         a = -g

         -g = - Aw² cos wt

          wt = cos⁻¹ (g / Aw²)

          t = 1 / w cos⁻¹ (g / Aw²)

angular velocity and frequency are related

        w = 2π f

        w = 2π 4

         w = 8π   rad / s

remember that the angles are in radians

          t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))

          t = 0.039789 1.3473

          t = 0.0536 s

let's find the speed for this time

          v = - A w sin wt

          v = - 0.07 8π sin (8π 0.0536)

          v = - 1,715 m / s

the distance is

           x = A cos wt

           x = 0.07 cos (8π 0.0536)

           x = 0.0156 m

Answer:

answer A is the correct one

Explanation:

Weber's law states that  "the smallest discernible change of a stimulus and proportional to the stimulus".

Applying this law to cases of optical intensity, the ratio must be

          k = cte = ΔI / I

where ΔI is the variation of the intensity and I is the value of the intensity

In general, for humans, the constant is 0.15 for the rods and 0.015 for the cones of the retina.

When reviewing the answers, answer A is the correct one, since in order for the previous relationship to be maintained, the magnitudes must rise proportionally