La distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distanciaLa distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distancia
Answer:
La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.
Explanation:
La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.
Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:
a ⇒ b
c ⇒ x
Entonces [tex]x=\frac{c*b}{a}[/tex]
La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?
1 km ⇒ 3280.84 pies
12 km ⇒ x
[tex]x=\frac{12 km*3280.84 pies}{1 km}[/tex]
x= 39370.08 pies
La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.
Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through?
index of reflection
index of refraction
angle of incidence
angle of reflection
Answer:
The index of refraction
What is the index of refraction?
The index of refraction tells us how much slower or faster light travels from one medium to another.
What is a medium? ⇒ the substance that light passes throughThe formula to find the refractive index of a substance is the following:
[tex]n=\dfrac{c}{v}[/tex]
[tex]c[/tex] = the speed of light in a vacuum ([tex]3.00*10^8[/tex] m/s)[tex]v[/tex] = the speed of light in the mediumThis makes the index of refraction defined as the ratio of:
the speed of light in a vacuum the speed of light in the material it passes throughAnswer:
index of refraction
Explanation:
Una esfera de aluminio de 250 g a una temperatura de 230 °C se introduce en 450 g de alcohol a una temperatura de 18 °C. Determine la temperatura de equilibrio del sistema. El calor específico del aluminio es 911 J kg-1 °C-1 y para el alcohol 2100 J kg-1 °C-1.
Answer:
La temperatura de equilibrio del sistema es de aproximadamente 59,17 °C
Explanation:
La masa de la esfera de aluminio, m₁ = 250 g
La temperatura inicial de la esfera de aluminio, T₁ = 230 °C
La capacidad calorífica específica del aluminio, c₁ = 911 J / (kg·°C)
La masa del alcohol, m₂ = 450 g
La temperatura inicial del alcohol, T₂ = 18 ° C
La capacidad calorífica específica del alcohol, c₂ = 2,100 J/(kg·°C)
Observamos que el punto de ebullición del alcohol ≈ 78,37°C
Sea 'T' la temperatura de equilibrio del sistema, obtenemos;
ΔQ₁ = ΔQ₂
ΔQ₁ = Calor transferido desde el aluminio = m₁ · c₁ · (T₁ - T)
ΔQ₂ = Calor transferido al alcohol = m₂ · c₂ · (T - T₂)
∴ m₁ · c₁ · (T₁ - T) = m₂ · c₂ · (T - T₂)
Por lo tanto, por sustitución, obtenemos
250 × 911 × (230 - T) = 450 × 2100 × (T - 18)
Usando una calculadora gráfica, tenemos;
T = 277570/4691 °C ≈ 59,17 °C
La temperatura de equilibrio del sistema, T ≈ 59.17 °C
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Newton's second law states that force is equal to
weight times acceleration
weight divided by mass
mass times weight
mass times acceleration
Answer:
mass times acceleration
Problem 31:
The parallel circuit depicts three resistors connected to a voltage source.
The voltage source (AVtot) is a 110.-V source and the resistor values are
15.4 QR1), 21.9 D (R2) and 11.7 Q (R3).
b. Determine the current in each branch resistor.
c. Determine the total current in the circuit.
Answer:
Bi. Current in 15.4 Ω (R₁) is 7.14 A.
Bii. Current in 21.9 Ω (R₂) is 5.02 A.
Biii. Current in 11.7 Ω (R₃) is 9.40 A.
C. Total current in the circuit is 21.56 A.
Explanation:
Bi. Determination of the current in 15.4 Ω (R₁)
Voltage (V) = 110 V
Resistance (R₁) = 15.4 Ω
Current (I₁) =?
V = I₁R₁
110 = I₁ × 15.4
Divide both side by 15.4
I₁ = 110 / 15.4
I₁ = 7.14 A
Therefore, the current in 15.4 Ω (R₁) is 7.14 A.
Bii. Determination of the current in 21.9 Ω (R₂)
Voltage (V) = 110 V
Resistance (R₂) = 21.9 Ω
Current (I₂) =?
V = I₂R₂
110 = I₂ × 21.9
Divide both side by 21.9
I₂ = 110 / 21.9
I₂ = 5.02 A
Therefore, the current in 21.9 Ω (R₂) is 5.02 A
Biii. Determination of the current in 11.7 Ω (R₃)
Voltage (V) = 110 V
Resistance (R₃) = 11.7 Ω
Current (I₃) =?
V = I₃R₃
110 = I₃ × 11.7
Divide both side by 11.7
I₃ = 110 / 11.7
I₃ = 9.40 A
Therefore, the current in 11.7 Ω (R₃) is 9.40 A.
C. Determination of the total current.
Current 1 (I₁) = 7.14 A
Current 2 (I₂) = 5.02 A
Current 3 (I₃) = 9.40 A
Total current (Iₜ) =?
Iₜ = I₁ + I₂ + I₃
Iₜ = 7.14 + 5.02 + 9.40
Iₜ = 21.56 A
Therefore, the total current in the circuit is 21.56 A
can someone help me :/ a 30 kg cart is accerlerating at 3 m/s2 (tiny 2) How much force is involved?
Explanation:
Write what you know
F = ma
m = 30kg
a = 3
Plug and chug
fun fact:numbers that are exponents are called superscripts
The energy principle states that:
Energy can be destroyed.
Energy can be created.
Energy cannot be created or destroyed.
Energy cannot be created, but it can be destroyed.
brainleist to correct answer
The energy principle states that:
[tex]\sf\purple{Energy \:cannot \:be \:created \:or\: destroyed.✅}[/tex]
Law of conservation of energy or the first law of thermodynamics states that energy can neither be created nor destroyed; it can only be transferred or changed from one form to another.[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.}}}}}[/tex]
HELP ASAP need it soon
Answer:
The answer to your question is given below.
Explanation:
The energy and frequency of a wave are related by the following equation:
E = hf
Where
E => is the energy.
h => is the Planck's constant.
f => is the frequency
From the formula i.e
E = hf
The energy is directly proportional to the frequency. Thus, an increase in the frequency of the wave will lead to an increase in the energy and a decrease in the frequency of the wave will lead to a decrease in the energy.
Considering the question given above, if the frequency of the wave is decreased, then the energy of the wave will also decrease.
Answer this Science question NO LINKS !!!!
A plane travels at an average speed of 600 kilometers per hour. How long
does it take the plane to travel 120 kilometers?
Answer:
12 mins
Explanation:
120km/600km = 1/5 or .2
60mins*0.2 = 12 mins
A tugboat can do 9,500,000 J of work through a distance of 35 m. What is the force exerted by the tugboat?
Answer:
271,428 4/7 N
Explanation:
Work done = force x distance.
9,500,000 = 35(force)
Force = 9,500,000 ÷ 35 = 271,428 4/7 N
Una rueda gira con una frecuencia de 530 rpm. Determina la velocidad angular, el periodo y la frecuencia.
Answer:
donde esta la bibliotekaaa
Explanation:
dfghj
Por definición de velocidad angular, periodo y frecuencia del M.C.U., la frecuencia, la velocidad angular y el periodo de la rueda son 8.833 Hz, 55.50 rad/s y 0.113 segundos respectivamente.
Movimiento Circular UniformeEl movimiento circular uniforme (M.C.U.) es un movimiento de trayectoria circular en el que la velocidad angular es constante. Esto implica que describe ángulos iguales en tiempos iguales.
En otras palabras, el M.C.U. describe el movimiento de un cuerpo con una rapidez constante y una trayectoria circular, cambiando en cada instante de dirección.
FrecuenciaLa frecuencia mide el número de revoluciones o vueltas completadas por unidad de tiempo. La frecuencia se expresa en s⁻¹ o Hertz (Hz).
Se define como:
f= ω÷ (2π) siendo ω la velocidad angular.
PeriodoEl periodo (T) es el tiempo que el cuerpo emplea en dar una vuelta completa; es decir que las características del movimiento son las mismas cada T segundos.
Se define como:
T= (2π)÷ ω siendo ω la velocidad angular.T= 1÷ f siendo f la frecuencia.Velocidad angular, periodo y frecuencia de una ruedaUna rueda gira con una frecuencia de 530 rpm. Expresada en las unidades correspondientes, la frecuencia de la rueda es:
f = 530rpm×[tex]\frac{1 min}{60 segundos}[/tex]
f = 8.833 Hz
Siendo la frecuencia f= ω÷ (2π), entonces la velocidad angular es calculada como:
ω= f× (2π)
Entonces, la velocidad angular de la rueda es:
ω= 8.833 Hz× (2π)
ω= 55.50 rad/s
Finalmente, por definición de periodo, el periodo de la rueda es:
T= 1÷ 8.833 Hz
T=0.113 segundos
En resumen, la frecuencia, la velocidad angular y el periodo de la rueda son 8.833 Hz, 55.50 rad/s y 0.113 segundos respectivamente.
Aprende más sobre el movimiento circular uniforme:
https://brainly.com/question/24081772
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The global positioning system (GPS) uses electromagnetic waves sent between computer devices on Earth and satellites in space to automatically determine the exact position of the computer device on Earth. In order to create this technology, engineers had to apply Einstein’s theory of relativity, which explains how electromagnetic waves change direction in space.
During which stage of developing the global positioning system did engineers first apply the theory of relativity to design the product?
first
second
third
fourth
Answer:
SECOND
Explanation:
I got it wrong when I put in First, and Edge told me the answer was b. SECOND
(GIVE THE ANSWER OR GET REPORTED) Please solve it with the steps
Answer:
4. a = 2 m/s² 5. m = 20 kg
Explanation:
4.
Mass of the book, m = 2kg
The net force acting on the book, F = 4 N
Let a is the acceleration of the book. We know that,
F = ma
Put all the values,
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{4}{2}\\\\a=2\ m/s^2[/tex]
So, the acceleration of the book is equal to [tex]2\ m/s^2[/tex].
5.
Kinetic energy of the car, K = 4000 J
Speed of the car, v = 20 m/s
We need to find the mass of the car. The formula for the kinetic enregy is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\m=\dfrac{2K}{v^2}\\\\m=\dfrac{2\times 4000}{(20)^2}\\\\m=20\ kg[/tex]
So, the mass of the car is equal to 20 kg.
Background Information: Energy can not be created or destroyed. Stored energy is called
potential energy, and the energy of motion is called kinetic energy. Due to gravity, potential
energy changes as the height of an object changes, this is called gravitational potential energy.
Objective: to determine the relationship between height and gravitational potential energy.
Problem: How does the drop height (gravitational potential energy) of a ball affect the bounce
height (kinetic energy) of the ball?
Hypothesis: If the gravitational potential energy (drop height) of a ball is increased, then the
kinetic energy (bounce height) will (increase/decrease/remain the same) because
Variables: Independent variable (known information) is
Dependent variable (unknown information) is
Constants (variables kept the same for accuracy) are
Materials: List all the materials used in this experiment.
Procedure: Follow the steps below to conduct your experiment. Be sure to record all data and
any observations during the experiment. Follow all safety rules.
1. Tape the meter stick to the side of the lab table with the 0-cm end at the bottom and the 100-cm end at the
top. Be sure that the meter stick is resting flat on the floor and is standing straight up.
2. Choose a ball type and record the ball type in the data table.
3. Use the triple beam balance to determine the mass of the ball and record the ball’s mass in the data table.
4. Calculate the gravitational potential energy (GPE) for the ball at each drop height. Record GPE in data table.
a. GPE = ball mass x drop height
5. For Trial 1, hold the ball at a height of 40 cm, drop the ball carefully and observe the bounce height. Record
the bounce height in the data table.
6. Drop the ball 4 more times from 40 cm, recording the bounce height each time, for a total of 5 drops.
7. For Trial 2, repeat steps 5 and 6 but drop the ball from a height of 50 cm. Record the 5 bounce heights in
the data table.
8. For Trial 3, drop the ball five times from 60 cm and record the 5 bounce heights in the data table.
9. For Trial 4, drop the ball five times from 70 cm and record the 5 bounce heights in the data table.
10. For Trial 5, drop the ball five times from 80 cm and record the 5 bounce heights in the data table.
11. For Trial 6, drop the ball five times from 90 cm and record the 5 bounce heights in the data table.
12. For Trial 7, drop the ball five times from 100 cm and record the 5 bounce heights in the data table.
13. Repeat steps 2 through 12 for a different type of ball.
14. Calculate the average bounce height of the 5 drops for each drop height. Record the average bounce height
in the data table. Calculate the average bounce height for all Trials.
a. To calculate average: Add the 5 bounce heights for a trial then divide the total by 5 drops. Example
for Trial 1: drop1 + drop2 + drop3 + drop4 +drop5 = total; total divided by 5 = average bounce height.
i. Sample: Trial 1: 5+6+5+5+7 = 28 28/5 = 5.6 average bounce height.
15. Plot the average bounce heights on a line graph. Place the independent variable of drop height on the x-axis
and place the dependent variable of bounce height on the y-axis. Label the line with the ball type.
16. CHALLENGE: Repeat all necessary steps for a third ball type and include this data on the graph.
17. Answer discussion questions and write your lab summary.
Data:
Ball Type: Ball Mass (g) = .
Gravitational
Potential Energy
(GPE)
Bounce Height (cm)
Drop
Height (cm)
GPE = mass x height Drop 1 Drop 2 Drop 3 Drop 4 Drop 5 Average
Bounce
Height
40
50
60
70
80
90
100
Ball Type: Ball Mass (g) = .
Gravitational
Potential Energy
(GPE)
Bounce Height (cm)
Drop
Height (cm)
GPE = mass x height Drop 1 Drop 2 Drop 3 Drop 4 Drop 5 Average
Bounce
Height
40
50
60
70
80
90
100
Discussion Questions: Answer using complete sentences.
1. Describe the relationship between drop height and the bounce height.
Was the relationship the same for both ball types that you tested?
2. Compare your gravitational potential energy to your bounce height for each trial. Describe
the relationship between GPE and bounce height.
3. Look at the results of both ball types you tested.
a. Which ball type had the most gravitational potential energy?
b. Which ball type has the most mass?
c. Describe the relationship between mass and GPE.
4. What are the variables that affect gravitational potential energy of an object?
Conclusion: Write a conclusion, using complete sentences, that states the following: if your
hypothesis was supported or negated; and what the real answer to the problem is.
Summary: Write a three paragraph summary using our standard format (1. what you were doing
and why you were doing it, 2. what you learned while doing this experiment, and 3. how what you
learned relates to your life).
Answer:
this question is very lengthy and even nit at all understandable so pls can u explain it in breif
Answer:
Explanation:
I did it and this is what i got
What is the pressure at a depth of 15 cm brine of density 1.2/cm³?
P = density × gravity acceleration × height
P = 1200 × 9.81 × 15/100
P = 1765.8
From the question we have got;
Height: 15 cm = 15/109m = 15 × 10-²Density: 1.2 cm³= 1.2g/cm³ = 1.2 × 10³ kg/m³Acceleration due to gravity:10m/s²Pressure = ρgh
= 15 × 10-² × 1.2 × 10³ × 10
= 1800 Nm-²
5. A 905 kg test car travels around a 3.04 km circular track. If the magnitude of the centripetal force is 2100 N. What is the car's speed?
Answer:
Explanation:
The equation for centripetal force is
[tex]F=\frac{mv^2}{r}[/tex]. We have all the values we need except for the radius. We have the circumference of the circle, though, so we will solve for the radius using that and the fact that C = 2πr:
3.04 = 2(3.1415)r and
r = .484 m, to the correct number of sig fig's.
Now that we have everything we need and isolating the v NOT squared:
[tex]v=\sqrt{\frac{rF}{m} }[/tex] and filling in:
[tex]v=\sqrt{\frac{(.484)(2100)}{905} }[/tex] . This answer will need 2 sig fig's since 2100 has 2 sig fig's in it. That means that the velocity of the test car is
1.1 m/sec
best explanation for why oceans have 2 different types of currents
Answer:
Hunubunununununununnunu
A power plant burns fuel to convert water to steam. During the process, 83% of the heat produced is transferred. Of the heat carried by the steam, 40% is converted to the mechanical energy of the spinning turbine. Which best describes the overall efficiency of the heat-to-work conversion in the plant
Answer: [tex]33.2\%[/tex]
Explanation:
Given
83% heat is transferred and out of this, only 40% is converted to the mechanical energy of spinning turbine.
The overall efficiency is the product of all the efficiency i.e.
[tex]\eta_t=0.83\times 0.4\\\eta_t=0.332\ \text{or}\ 33.2\%[/tex]
What is the voltage produced by a 180 amp current flowing through a resistance of 0.5 ohms?????
Answer:
90
Explanation:
v=ir
voltage,I current,while r is the resistance
v=180×0.5
90
Which term BEST describes energy
stored in batteries and food?
potential energy
kinetic energy
heat energy
chemical energy
The voltage drop across any component is
the difference in voltage referenced to
between the two terminals of
the component.
Answer:
The voltage si the nevery Hokage de Naruto y fue su amigo
An object being pulled towards Earth by gravity is said to be in __.
Answer:
in orbit
Explanation:
An object in orbit, like an object falling to the ground, is pulled toward Earth's center. If the object moves far enough forward as it falls, it orbits around Earth instead of hitting the ground.
An artificial sattelite is moving in a circular orbit of radius 42250km calculate its speed if it takes 24 hours to revolve around the earrth
Answer:
v = 11061.02 km/h
Explanation:
Given that,
The radius of the circular orbit, r = 42250 km
Time taken to revolve around the Earth, t = 24 hours
We need to find the speed of the satellite. We know that, speed of an object is equal to total distance traveled divided by the time taken. So,
[tex]v=\dfrac{2\pi r}{T}[/tex]
Put all the values,
[tex]v=\dfrac{2\pi \times 42250}{24}\\\\v=11061.02\ km/h[/tex]
So, the speed of the satellite is equal to 11061.02 km/h.
Can any one help plz in the question ?
Yes, No, yes
mark me brainliestttt :)))
Answer:
yes
no
no
I think Chek the answer
why doesn't the arrow fly forever
Because gravity pulls it down
Parallel Circuits:
A) are rarely used in the wiring in homes
B) always need more voltage than series circuits
C) will have positive charges flowing in one branch, negative charges in the other
D) provide more than one path fo current flow
Answer:
D.)
Explanation:
the current separates on each branch according to the resistance it experience.
Answer:D
Explanation:
Problem 6. A negatively charged particle is placed in a uniform electric field directed
from South to North. In which direction will the particle move after it is released?
A) West
B) East
C) South
D) North
E) North-West
po
UP
A
B
Meeting na
D
Activate!
Go to Setting
C) South
As we know that, north is considered as negative, and south as positive. Now, the charge on particle is negative; so, when we will release it, it will move towards the south.
In the scenario where the negatively charged particle is placed in a uniform electric field, the direction will be C. South.
What is an electric field?It should be noted that an electric field simply means the physical field which surrounds electrically charged particles.
Since the negatively charged particle is placed in a uniform electric field directed from South to North, the direction that the particle move after it is released is South.
Learn more about electric field on:
https://brainly.com/question/14372859
Compared to yesterday, you did three times the work in one-third the time. To do so, your power output must have been ___
Answer: 9 times yesterday's power output
Explanation:
Power is the rate at which a certain work is done. The unit of power is Watt and it's equal to work divided by the time taken to complete the work.
In this scenario, since the person did three times the work in one-third the time, then the power output will be:
= Work done / Time
= 3/⅓
= 3 × 3
= 9
Therefore, the power output must have been 9 times yesterday's power output.
sphere A is charged positively where sphere B is neutral.the sphere are brought near to each other
A) there is no electric force between the sphere
B) there is repulsive electrostatic force between the sphere
C)there is an attractive electrostatic force between the sphere
D)there is repulsive electrostatic force applied only on sphere B
D)there is repulsive electrostatic force applied only on sphere B