Verificar
El gráfico de la velocidad de una bola que rueda por un canal se muestra en la
figura:

a) La velocidad de la bola luego de 5 s.
0.6m/s
b) La distancia recorrida en el intervalo t : (0-5) s
1.5m
c) La distancia recorrida en el intervalo t: (5-10) s
3m
d) La aceleración de la bola en el intervalo t: (10-12) s.
-0.4m/s^2
e) La distancia total recorrida.
6.9m
f) La rapidez media de la bola
0.58m/s

VerificarEl Grfico De La Velocidad De Una Bola Que Rueda Por Un Canal Se Muestra En La Figura:a) La Velocidad

Answers

Answer 1

El análisis de un grafico de velocidad versus tiempo permite encontrar los resultados para las preguntas son:

a) La velocidad es v= 0,6 m/s

b) La distancia recorrida  dₐ = 1,5 m

c) La distancia recorrida entre t= 5 y 10 s es  \(d_b\) = 3,0 m

d) La aceleración   a = -0,4 m/s²

e) La distancia total recorrida  es \(d_{total}\) = 5,3 m

f) La velocidad media es  \(v_{media}\) = 0,44 m/s

La cinemática estudia el movimiento de los cuerpos, buscando relaciones entre la posición, la velocidad y la aceleración de los cuerpos.

La velocidad es definida como el cambio de la posición en función del tiempo y la aceleración es el cambio de la velocidad en el intervalo de tiempo.

           .\(v= \frac{\Delta x}{t} \\a = \frac{\Delta v}{t}\)

           

Donde v y a son la velocidad y aceleración, respectivamente, t es el tiempo y Δx y Δv son la variación de la posición y la velocidad, respectivamente.  

En el adjunto muestran un grafico de la velocidad en función del tiempo, donde del área bajo la curva obtenemos la distancia recorrida y la pendiente corresponde a la aceleración del cuerpo.

Respondamos las preguntas:

a)  La velocidad de la bola a los 5 s.

Como es un grafico de velocidad versus tiempo, la velocidad puede ser leída directamente, en el grafico tenemos un valor de

          v= 0,6 m/s  

b) La distancia recorrida corresponde al área bajo la curva.

         \(d_a = \frac{v_f - v_o}{2} t \\d_a = \frac{0.6-0}{ 2}\ 5\)  

         d = 1,5 m

c) La distancia en el intervalo de 5 a 10 s.

En este intervalo la velocidad es constante, el área es

        \(d_b = v \Delta t\\d_b = 0.6 \ 5\\\)  

        d =  3 m

d) La aceleración en el intervalo  10 a 12 s.

         

La aceleracion es la pendiente del grafico

        a = \(\frac{\Delta v}{\Delta t}\)  

        a = \(\frac{-0.2 - 0.6}{12-10}\)

        a = -0,4 m/s²

El signo negativo indica que la aceleración se opone a la velocidad del cuerpo.

e) La distancia total recorrida  

       \(d_{total} = d_a + d_b + d_c\)

      dₐ = 1,5 m

      \(d_b\) = 3,0 m

Busquemos la distancia en el ultimo intervalo

         \(d_c = \frac{0.6 + 0.2}{2} ( 12-10)\)

       \(d_c\) = 0,8 m

Substituimos

     \(d_{total}\) = 1,5 + 3,0 + 0,8

     \(d_{total}\)  =  5,3 m

f) La rapidez  de la bola

     

La rapidez prmedio es la relación entre la distancia recorrida en el tiempo.

         \(v_{avg} = \frac{\Delta x}{\Delta t }\)  

         \(v_{avg} = \frac{5.3}{12}\)

         \(v_{avg}\)  =0,44 m/s

En conclusión usando el análisis de un grafico de velocidad versus tiempo podemos encontrar los resultados para las preguntas son:

         a) La velocidad es v= 0,6 m/s

         b) La distancia recorrida  da = 1,5 m

         c) La distancia recorrida entre t= 5 y 10 s es  db = 3,0 m

         d) La aceleración   a = -0,4 m/s²

         e)  La distancia total recorrida  es \(d_{total}\) = 5,3 m

          f) La velocidad media es  \(v_{avg}\) = 0,44 m/s

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Related Questions

Why are weathering, erosion and deposition a NECESSARY process in the rock cycle?
ANSWER THIS NOW PLEASE! AND YOU GET 225 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answers

Answer:

YESS well it is partly nessary but it depends on the situation

Explanation:

Answer:

Weathering, erosion, and deposistion are necessary processes in the rock cycle because:

Explanation:

First, start with igneous rocks. magma erupts (Extrusive igneous rocks) or solidifies in the sub-surface of the earth (Intrusive igneous rock). when they are exposed Weathering and erosion occur which is a slow breakdown of rock through the wind, water, or other processes. The weathered pieces (sediments) move to other places by wind or water and get deposited someplace else. When there are enough sediments and there is overburden pressure on these sediments, they become a sedimentary rock. Due to overburden pressure, they become metamorphic rocks. now the thing to understand here is that when metamorphic rocks are exposed, they too undergo weathering and erosion and their pieces also become sedimentary rocks.

A carbon ion with 10 electrons and 6 protons would have a charge of ?

Answers

Answer:

-4

Explanation:

Electrons are negative while protons are positive.

To find the charge, add the number of protons to the number of electrons.

6+(-10)

6-10

-4 is the charge

Find distance between two object of radius 6 cm and 2 cm

Answers

The distance between two objects of radius 6 cm and 2 cm is zero

To find the distance between two objects with radii of 6 cm and 2 cm, we need to consider the center-to-center distance between the objects and subtract the sum of their radii.

Let's denote the radii of the objects as r1 = 6 cm and r2 = 2 cm.

The distance between the centers of the objects can be represented as d = r1 + r2. Adding the radii ensures that we account for the space occupied by both objects.

Substituting the values, we have d = 6 cm + 2 cm = 8 cm.

Now, to find the actual distance between the objects, we subtract the sum of their radii from the center-to-center distance:

Distance = d - (r1 + r2) = 8 cm - (6 cm + 2 cm) = 8 cm - 8 cm = 0 cm.

The resulting distance is 0 cm, indicating that the objects are in direct contact with each other. This means that their surfaces are touching. When the distance between two objects is zero, it implies that they are overlapping or in physical contact. In this case, since the distance is equal to 0 cm, the two objects are touching each other, with their surfaces coming into contact.

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A motor with an operating resistance of 30.0 ohms is connected to a voltage source 4.0 amps of current flow in the circuit. What is the voltage of the source

Answers

Answer:

120 V.

Explanation:

voltage V=IxR

= 4x30

= 120 V.

explain the main development stages of the tropical hurricanes/cyclones and discuss their impacts according to the Saffir Simpson Hurricane Wind Scale​

Answers

Answer:

The Saffir-Simpson Hurricane Wind Scale is a 1 to 5 rating based only on a hurricane's maximum sustained wind speed. This scale does not take into account other potentially deadly hazards such as storm surge, rainfall flooding, and tornadoes.

The Saffir-Simpson Hurricane Wind Scale estimates potential property damage. While all hurricanes produce life-threatening winds, hurricanes rated Category 3 and higher are known as major hurricanes*. Major hurricanes can cause devastating to catastrophic wind damage and significant loss of life simply due to the strength of their winds. Hurricanes of all categories can produce deadly storm surge, rain-induced floods, and tornadoes. These hazards require people to take protective action, including evacuating from areas vulnerable to storm surge.

Explanation:

pls mark as branlist.

Tropical hurricanes or cyclones are usually destructive with great wind speeds of 3 and above according to the Saffir Simpson Hurricane Wind Scale with huge amounts of rainfall resulting in flooding.

What is the Saffir Simpson Hurricane Wind Scale?

The Saffir Simpson Hurricane Wind Scale is scale that classifies wind based on wind speed using a scale of 1-5.

This scale does not take into account wind surge, amount of rainfall and other characteristics of storms.

Storms rated 3 and above are particularly strong and destructive.

Those rated below 3 are mild and less destructive.

Tropical hurricanes or cyclones are usually destructive with great wind speeds of 3 and above according to the Saffir Simpson Hurricane Wind Scale with huge amounts of rainfall resulting in flooding.

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importance of projectile​

Answers

The importance of that is people are smart and so r u so do this question and so will u

If the average electrical power usage of an American home is 2 kW, then for how
many homes could electricity needs be met by one 2 MW wind turbine?

Answers

1,000 homes when it's spinning.

No homes when it's not spinning.

The number of homes that the wind turbine can carry is 1000 homes

Power is the ratio of work done by an object with respect to time. The unit of power is in Watts (W).

Given the average electrical power usage of an American home to be 2 kW

Since 1kW = 100W

Average power per American home = 2000W = 2 * 10³Watts

Average power of a turbine = 2MW = 2 * 10⁶Watts

Number of homes that the wind turbine can carry = \(\dfrac{2 \times 10^6}{2 \times 10^3}\)

Number of homes that the wind turbine can carry = \(\dfrac{10^6}{10^3}\)

This shows that the number of homes that the wind turbine can carry is 1000 homes

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Which of the following is a vector quantity
weight
temperature
acceleration
distance

Answers

Answer:

weight, acceleration

Explanation:

weight = mass x gravity(meaning the direction of the mass)

acceleration = v-u/t

v-u is the change in velocity

(Briefly explain what would happen, if there is no existence of capacitor in our world?​

Answers

If there is no existence of capacitors in our world there would be no electrical or electronic engineering.

A capacitor is a device that stores electrical energy in an electric field. It has two terminals and is a passive electrical component. Capacitance refers to a capacitor's effect. A capacitor commonly referred to as a condenser is one of the fundamental parts needed to create electronic circuits. Without fundamental parts like resistors, inductors, diodes, transistors, etc., a circuit's design is incomplete or it won't work properly.

Energy storage is capacitors' most popular application. Power conditioning, signal coupling or decoupling, electronic noise filtering, and remote sensing are further applications. Capacitors are employed in a wide variety of industries and have integrated into daily life due to their numerous applications.

There are numerous significant uses for capacitors. They are employed in digital circuits, for instance, to prevent the loss of data saved in big computer memories during a brief loss of power. The electric energy held in such capacitors keeps the data from being lost during a brief power outage.

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A capacitor is formed from two square plates of edge length a and separation d, with d <

Answers

In an electrical circuit, a capacitor is a component that is used to hold charges.

A capacitor operates under the principle that when an earthed conductor is moved close to a conductor, its capacitance increases noticeably. As a result, a capacitor contains two plates with equal and opposing charges that are spaced apart.

Two metal plates, a substance known as a dielectric, and other components make up a capacitor.

The dielectric is positioned between the parallel metal plates, which are positioned very near to one another, to prevent them from touching.

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if the body is floating in a liquid then can we say that the rise in the level of the liquid is equal to the height of the body

Answers

Yes, if a body is floating in a liquid, the rise in the liquid level is equal to the body height. This phenomenon is known as Archimedes' principle.

Archimedes' principle says when a body is immersed in a fluid (liquid or gas), it experiences an upward buoyant force equal to the weight of the fluid displaced by the body. Buoyant forces act in the opposite direction to gravity.

When a body floats in a liquid, it displaces a volume of liquid equal to its volume. As a result, the liquid level rises by an amount equal to the height of the submerged part of the body.

This principle holds for objects that float or are partially immersed in a liquid, such as a buoyant boat or a floating object. However, if the body sinks completely into the liquid, the liquid level rise will no longer be equal to its height. Instead, it depends on the density and volume of the submerged object.

What happens to the brightness of the lightbulb when its resistance is increased? *

Answers

Answer:

It will be dimmer than before

Answer:For parallel connection,the brightness would be dimmer, while for series connection it would be brighter

Explanation:

For parallel connection,resistance and brightness are inversely proportional.meaning as resistance increases, brightness decreases.

For series connection,resistance and brightness are directly proportional. Meaning as the resistance increases, brightness also increases.

What is the solution?

What is the solution?

Answers

Answer:

1) x = 30 - 8 t

2) x = -10

3) x = -10 + 5 t

4) x = -10 - 4 t

Explanation:

Motion 1:  constant negative velocity calculated via the points (0, 30) and (5, -10) rendering the equation of motion  x = 30 - 8 t

Motion 2: constant position over time, so the object is not moving, and the equation of motion is x = -10

Motion 3: constant positive velocity estimated via the points (0, -10) and (2, 0), and the equation of motion is:  x = -10 + 5 t

Motion 4: constant negative velocity estimated via the points (0, -10) and (5, -30), and the equation of motion is:  x = -10 - 4 t

The starting position of motion 1 is 30 meters

the starting position for the other 3 motions is - 10 meters. And none of them is accelerated (acceleration = zero for all).

So long as there are(is) ______ there is no movement.

A. friction
B. unbalanced forces
C. balanced forces
D. forces ​

Answers

Answer:

answer is friction. MCQ A is answer

how are series and parallel circuits different in terms of current and voltage? how are they similar? why?

Answers

Answer:

i got u

Explanation:

In terms of current, series and parallel circuits behave differently. In a series circuit, the current is the same at all points in the circuit. This is because the current has only one path to flow through and is restricted by the resistance of the components in the circuit. In contrast, in a parallel circuit, the current is divided between the branches of the circuit and can vary at different points.

In terms of voltage, series and parallel circuits also differ. In a series circuit, the voltage is divided among the components of the circuit according to their resistance. The total voltage of the circuit is the sum of the voltages across each component. In a parallel circuit, the voltage is the same across all the components. This is because each component is connected directly to the voltage source.

One similarity between series and parallel circuits is that they both obey Ohm's law, which states that the current in a circuit is directly proportional to the voltage and inversely proportional to the resistance. Another similarity is that both types of circuits can contain various types of components, such as resistors, capacitors, and inductors.

The reason why series and parallel circuits behave differently in terms of current and voltage is due to the different ways in which the components are connected. In a series circuit, the components are connected end to end, forming a single path for current to flow. In a parallel circuit, the components are connected side by side, providing multiple paths for current to flow. This difference in the arrangement of the components affects how current and voltage are distributed throughout the circuit.

A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds upward. If the ball hits the floor with a speed of 2.60 m/s, rebounds with a speed of 1.84 m/s, and is in contact with the floor for 1.40 ms,determine the following.
(a) magnitude of the change in the ball's momentum (Let up be in the positive j hat direction.)
_______________kg

Answers

Answer:

-0.209 kg.m/s

Explanation:

The mass of the ball, m = 275g or 0.275 kg

Speed or velocity, v = 2.60 m/s

Momentum, P = mv

Momentum when velocity is 2.60 = 0.275 x 2.60 = 0.715 kg.m/s

Speed or velocity, v = 1.84 m/s

Momentum, P = mv

Momentum when velocity is 1.84= 0.275 x 1.84 = 0.506 kg.m/s

Change in magnitude =  0.506 - 0.715 = -0.209 kg.m/s

In the figure below, what is the maximum speed of a 2.0 g particle that oscillates between x = 2.0 mm and x = 8.0 mm?

In the figure below, what is the maximum speed of a 2.0 g particle that oscillates between x = 2.0 mm

Answers

In the figure below, the maximum speed of a 2.0 g particle that oscillates between x = 2.0 mm and x = 8.0 mm is 63 m/s.

What is oscillation?

Periodic or oscillatory motion is defined as a motion that repeats itself. Due to a restoring force or torque, an object in such motion oscillates around an equilibrium position.

Energy is conserved in oscillation.

So, the maximum loss in Potential Energy = Maximum gain in Kinetic energy

5 J - 1 J = Maximum gain in Kinetic energy

Maximum gain in Kinetic energy = 4 J

\(0.5 \times m \times vmax^2 = 4\\0.5 \times 2.0 \times 10^-3\; Kg \times vmax^2 = 4 J\\vmax^2 = 4 \times 10^3 m^2/s^2\\vmax = 63.2 m/s\)

Therefore, the maximum speed of a 2.0 g particle that oscillates between x = 2.0 mm and x = 8.0 mm is 63 m/s.

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The known values are given in the table below them. Find the missing values.

The known values are given in the table below them. Find the missing values.

Answers

The value of  the missing resistance, R₃ = 10.35 ohms.

The value of the missing voltages, V₁ = 6 V, V ₃ = 24 V.

The value of the missing currents, I₁ = 3 A, I₃ = 2.32 A.

What is the value of missing component of the circuit?

The values of the missing component of the circuit is calculated by applying the following formula.

The total resistance of the circuit;

For R₂, R₃, 1/R = 1/R₂ + 1/R₃

1/R = 1/12 +  1/R₃

1/R = (R₃ + 1)/(12R₃)

R = 12R₃ / (R₃ + 1)

For, R₁, R₂ and R₃, total resistance;

R = 12R₃ / (R₃ + 1) + R₁

R = [12R₃ / (R₃ + 1)] + 2

R = (12R₃ + 2(R₃ + 1) ) / (R₃ + 1)

R = (12R₃ + 2R₃ + 2 ) / (R₃ + 1)

R = (14R₃ + 2 ) / (R₃ + 1)

The total current in circuit is calculated as;

I = V/R

I = 30 / R

I = ( 30 ) / (14R₃ + 2 ) / (R₃ + 1)

I = (30R₃ + 30) / (14R₃ + 2) ------- (1)

The voltage in parallel circuit is the same

V₂ = V₃ = 24 V

V₃ = IR₃

24 = IR₃

I = 24/R₃  --------- (2)

Solve (1) and (2) together as follows;

24/R₃ = (30R₃ + 30) / (14R₃ + 2)

30R₃² - 306R - 48 = 0

Solve the quadratic equation, using formula method.

R₃ = 10.35 ohms

I₃ = V₃/R₃

I₃ = 24 V / 10.35

I₃ = 2.32 A

If the voltage drop at R₂ and R₃ = 24 V, the voltage drop at R₁ = 30V - 24 V = 6 V

The current in R₁ = V₁/R₁ = 6 V / 2 V = 3 A

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A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 7.2 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall

Answers

Answer:

480N

Explanation:

Given data

M1= 1.2kg

U1= 8m/s

V1=7.2 m/s

T= 2ms= 0.002

We know that the impulse expression is

P=FT

M(U1-V1)= FT

substitute

1.2(8-7.2)= F*0.002

1.2*0.8=  F*0.002

0.96=  F*0.002

F= 0.96/0.002

F=480N

Hence the force is 480N

A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50×10−5T and points into the Earth at an angle of 58.0 below a line pointing due north. A 6.90-A clockwise current passes through the coil.

Answers

The Earth's magnetic field at the location has a magnitude of 5.50×10^−5 T and points into the Earth at an angle of 58.0 degrees below a line pointing due north.

What is Magnetic Field?

A magnetic field is a region of space surrounding a magnet or a current-carrying conductor in which magnetic forces are exerted on other magnetic objects or moving charged particles. Magnetic fields are characterized by their direction, magnitude, and polarity. The direction of a magnetic field is defined as the direction in which a magnetic north pole would be pulled or aligned, and is conventionally represented by magnetic field lines that form closed loops.

Based on the information provided, it seems like you have described a situation where a circular coil with a diameter of 14.0 cm and containing nine loops is lying flat on the ground.

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A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.00×10−5T and points into the Earth at an angle of 58.0 ∘ below a line pointing due north. A 6.90-A clockwise current passes through the coil. Determine the torque on the coil, and which edge of the coil rises up: north, east, south, or west?

Question 4
1 pts
Which of the following is not evidence that kinetic energy has been
lost in a collision?
At least one of the objects is deformed after the collision.
At least one of the objects increases in temperature as a result of the collision.
The collision produces a sound.
One of the objects is at rest after the collision.
<

Answers

Answer:

one of the object is at rest after the collision

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?

Answers

Answer:

The greater of the two currents is 0.692 A

Explanation:

Given;

distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m

let the current in the first wire = I₁

then, the current in the second wire = 2I₁

length of the wires, L = 3.0 m

magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N

The magnitude of force on the two parallel wires is given by;

\(F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A\)

the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A

Therefore, the greater of the two currents is 0.692 A

A red ball is dropped from rest at a height of 7.80 m. A blue ball at a height of 11.7 m is thrown down at the same instant at 9.00 m/s. How long does it take the blue ball to catch up with the red ball?

Answers

Answer:

It takes the blue ball 0.4333 seconds to catch up with the red ball.

Explanation:

We can use the equation

\(y=v_ot-\frac{1}{2}gt^2\)

Note \(y\) is the change in position on the y-axis. So there is an initial and final component.

We can use that equation to find out at what time \(t\), the 2 balls will have traveled the same distance.

Let the red ball be object 1 and let the blue ball be object 2.

\(y_{o1} -(v_{o1}t-\frac{1}{2}gt^2)=y_{o2} -(v_{o1}t-\frac{1}{2}gt^2)\)

Solving for \(t\) gives us

\(t=-\frac{y_{o2} -y_{o1}}{v_{o1} -v_{o2}}\)

We are given

\(y_{o1}=7.8\)

\(y_{o2}=11.7\)

\(v_{o1}=0\)

\(v_{o2}=9\)

Substituting our values into the equation gives us

\(t=-\frac{11.7-7.8}{0-9}\)

\(t=-\frac{3.9}{0-9}\)

\(t=-\frac{3.9}{-9}\)

\(t=0.4333\)

what is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)
please someone help me

Answers

Answer:

60mph=26.8224meters per second

Explanation:

A 100 pF parallel-plate capacitor is made of two metal plates. The area of each plate is 100 mm2. What is the separation between the plates?Group of answer choices9.06x10-6 m8.85x10-6 m6.44x10-6 m5.60x10-6 m1.13x10-6 m

Answers

The capacitance C of a parallel-plate capacitor whose plates have an area A and are separated by a distance d is:

\(C=\frac{\varepsilon_0A}{d}\)

Where ε₀ is the vacuum permitivity:

\(\varepsilon_0=8.854\times10^{-12}\frac{F}{m}\)

Notice that the capacitance of the parallel-plate capacitor is given, as well as the area of its plates. Then, isolate d from the equation:

\(\Rightarrow d=\frac{\varepsilon_0A}{C}\)

Replace A=100mm^2, C=100pF and the value of vacuum permitivity to find the separation between the plates:

\(d=\frac{(8.854\times10^{-12}\frac{F}{m})(100\times10^{-6}m^2)}{(100\times10^{-12}F)}=8.854\times10^{-6}m\)

Therefore, the correct choice is: 8.85x10^-6m.

Find the locations of the object when its velocity is one-third of the maximum speed. Treat the equilibrium position as zero, positions to the right as positive, and positions to the left as negative.

Answers

Answer:

let x = A sin ω t        describes location of object at time t

V = A ω cos ω t        velocity of object at time t

Vmax = A ω       max speed occurs when ω t = 0      velocity / max velocity

V / Vmax = 1/3  = cos ω t

θ = ω t       angle being considered

cos θ = 1/3

1/3 max speed occurs at cos θ = ±1/3

θ = ±70.5 deg

10. Liquids are poured into a container and separate. They form a green layer on
the bottom, a blue layer in the middle, and a red layer on the top. What's true about
the density of the liquids?
A. The red liquid is the least dense.
B. The green liquid is the least dense.
C. The blue liquid is the least dense.
D. They all have the same density.

Answers

Option A is true: the red liquid is the least dense.

What happens when different liquids are poured into a container?

When different liquids are poured into a container, they will typically separate based on their densities. Liquids with higher densities will sink to the bottom, while liquids with lower densities will float on top.

In the scenario described in the question, we are told that the liquids have separated into three distinct layers: green on the bottom, blue in the middle, and red on top. Based on this layering, we can infer that the red liquid must be the least dense, since it is floating on top of the other two liquids.

Similarly, we can infer that the blue liquid is denser than the red liquid, but less dense than the green liquid, since it is sandwiched between them. And we can infer that the green liquid is the most dense, since it is at the bottom.

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Which best describes the way a sound wave is sent through the radio

Answers

Answer:

sound wave-electrical wave-radio wave

Answer:

D

Explanation:

on edg

a mass of 7.5kg has a weight of 30n on a certain planet calculate the acceleration due to gravitt on this planet

Answers

Answer:

Acceleration due to gravity on a certain planet = 4 m/s²

Explanation:

According to the question,

Weight = 30 N

Mass = 7.5 kg

Let acceleration due to gravity be 'a'

Formula:

Weight = Mass × Acceleration due to gravity

30 = 7.5 × a

a = 30/7.5

a = 4 m/s²

A wire is 24.5 m long at 4.00°C and is 1.40 cm longer at 34.0°C. Find the wire's coefficient of linear expansion (in (°C)−1)

Answers

Answer:

1.9×10⁻⁵ °C⁻¹

Explanation:

Coefficient of linear expansion: This can be defined as as increase in length of a material, per unit length per degree rise in temperature, The S.I unit is (K⁻¹).

From the question above,

α = ΔL/L(T₂-T₁)................... Equation 1

Where α = coefficient of linear expansion, ΔL = Change in length, L = original length, T₁ = Initial Temperature, T₂ = Final temperature.

Given: ΔL = 1.4 cm = 0.014 m, L = 24.5 m, T₁ = 4.00°C, T₂ = 34°C

Substitute these values into equation 1

α = 0.014/[24.5(34-4)]

α = 0.014/735

α = 1.9×10⁻⁵ °C⁻¹

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